text stringlengths 235 3.08k |
|---|
ribute 12x + 8 = 6x Move variables to one side 12x Subtract 12x from both sides 12x 6x Divide both sides by β Our Solution 269 Example 362. 2x 3 β 7x + 4 2 5 3) = 2(7x + 4) Distribute = Calculate cross product 15 = 14x + 8 Move variables to one side Subtract 10x from both sides Subtract 8 from both sides 5(2x β 10x 10x β β β 10x 15 = 4x + 8 8 8 β β β 4 23 = 4x Divide both sides by 4 4 23 4 = x Our Solution β β As we solve proportions we may end up with a quadratic that we will have to solve. We can solve this quadratic in the same way we solved quadratics in the past, either factoring, completing the square or the quadratic formula. As with solving quadratics before, we will generally end up with two solutions. Example 363) = (8)(3) β Calculate cross product FOIL and multiply (k + 3)(k β k2 + k β k2 + k (k + 6)(k k + 6 = 0 or k 6 β k = β 6 β β 24 6 = 24 Make equation equal zero 24 β 30 = 0 5) = 0 5 = 0 Subtract 24 from both sides Factor Set each factor equal to zero Solve each equation β β β + 5 = 5 Add or subtract 6 or k = 5 Our Solutions Proportions are very useful in how they can be used in many diο¬erent types of applications. We can use them to compare diο¬erent quantities and make conclusions about how quantities are related. As we set up these problems it is important to remember to stay organized, if we are comparing dogs and cats, and the number of dogs is in the numerator of the ο¬rst fraction, then the numerator of the second fraction should also refer to the dogs. This consistency of the numerator and denominator is essential in setting up our proportions. Example 364. A six foot tall man casts a shadow that is 3.5 feet long. If the shadow of a ο¬ag pole is 8 feet long, how tall is the ο¬ag pole? shadow height We will put shadows in numerator, heights in denomintor 270 3.5 6 8 x The man has a shadow of 3.5 |
feet and a height of 6 feet The ο¬agpole has a shadow of 8 feet, but we donβ²t know the height 3.5 6 8 x 3.5x = (8)(6) Multiply = This gives us our proportion, calculate cross product 3.5x = 48 Divide both sides by 3.5 3.5 3.5 x = 13.7ft Our Solution Example 365. In a basketball game, the home team was down by 9 points at the end of the game. They only scored 6 points for every 7 points the visiting team scored. What was the ο¬nal score of the game? home visiter x β x 9 6 7 We will put home in numerator, visitor in denominator Donβ²t know visitor score, but home is 9 points less Home team scored 6 for every 7 the visitor scored This gives our proportion, calculate the cross product β β 7(x 7x 7x β x 9 6 7 = β x 9) = 6x Distribute 63 = 6x Move variables to one side Subtract 7x from both sides 7x x Divide both sides by 1 1 β β 63 = β 1 β β β 63 = x We used x for the visitor score. 9 = 54 Subtract 9 to get the home score 63 β 63 to 54 Our Solution 271 7.6 Practice - Proportions Solve each proportion. 1) 10 a = 6 8 3) 7 6 = 2 k 5) 6 x = 8 2 7) m 1 5 = 8 β 2 2) 7 9 = n 6 4) 8 x = 4 8 6) n β 10 8 = 9 3 8) 8 5 = 3 x β 8 9) 2 9 = 10 p β 4 10) 9 n + 2 = 3 9 11) b β 10 7 = b 4 13) x 5 = x + 2 9 15) 3 10 = a a + 2 17 19) 7 1 = 4 x x β 21) x + 5 5 = 6 x β β 12) 9 4 = r r β 4 14) n 8 = n β 3 4 16 18) n + 8 10 = n β 4 9 20 22 23) m + 3 4 = 11 m β 4 24 25 26) 5 n + 1 = n β 10 4 27 28 29 30 Answer each question. Round your answer to the nearest tenth. Round dollar amounts to the nearest cent. 31) The currency in Western Samoa is the Tala. The |
exchange rate is approximately S0.70 to 1 Tala. At this rate, how many dollars would you get if you exchanged 13.3 Tala? 32) If you can buy one plantain for S0.49 then how many can you buy with S7.84? 272 33) Kali reduced the size of a painting to a height of 1.3 in. What is the new width if it was originally 5.2 in. tall and 10 in. wide? 34) A model train has a scale of 1.2 in : 2.9 ft. If the model train is 5 in tall then how tall is the real train? 35) A bird bath that is 5.3 ft tall casts a shadow that is 25.4 ft long. Find the length of the shadow that a 8.2 ft adult elephant casts. 36) Victoria and Georgetown are 36.2 mi from each other. How far apart would the cities be on a map that has a scale of 0.9 in : 10.5 mi? 37) The Vikings led the Timberwolves by 19 points at the half. If the Vikings scored 3 points for every 2 points the Timberwolves scored, what was the half time score? 38) Sarah worked 10 more hours than Josh. If Sarah worked 7 hr for every 2 hr Josh worked, how long did they each work? 39) Computer Services Inc. charges S8 more for a repair than Low Cost Computer Repair. If the ratio of the costs is 3 : 6, what will it cost for the repair at Low Cost Computer Repair? 40) Kelseyβs commute is 15 minutes longer than Christinaβs. If Christina drives 12 minutes for every 17 minutes Kelsey drives, how long is each commute? 273 7.7 Rational Expressions - Solving Rational Equations Objective: Solve rational equations by identifying and multiplying by the least common denominator. When solving equations that are made up of rational expressions we will solve them using the same strategy we used to solve linear equations with fractions. When we solved problems like the next example, we cleared the fraction by multiplying by the least common denominator (LCD) Example 366. 2 3 x 2(12) 3 x β 5(12) 6 5 6 = 3 4 3(12) 4 β = Multiply each term by LCD, 12 Reduce fractions 2(4)x β 5(2) = 3(3) Multiply 8x 10 = 9 Solve β + 10 + 10 Add 10 to |
both sides 8x = 19 Divide both sides by 8 8 x = Our Solution 8 19 8 We will use the same process to solve rational equations, the only diο¬erence is our 274 LCD will be more involved. We will also have to be aware of domain issues. If our LCD equals zero, the solution is undeο¬ned. We will always check our solutions in the LCD as we may have to remove a solution from our solution set. Example 367. 5x + 5 x + 2 + 3x = x2 x + 2 Multiply each term by LCD, (x + 2) (5x + 5)(x + 2) x + 2 + 3x(x + 2) = x2(x + 2) x + 2 Reduce fractions 5x + 5 + 3x(x + 2) = x2 Distribute 5x + 5 + 3x2 + 6x = x2 Combine like terms 3x2 + 11x + 5 = x2 Make equation equal zero Subtract x2 from both sides Factor Set each factor equal to zero Solve each equation β β x2 x2 2x2 + 11x + 5 = 0 (2x + 1)(x + 5) = 0 2x + 1 = 0 or x + 5 = 0 5 5 5 1 or x = β 1 β β 1 β β 2x = or β 5 + 2 = 1 2 or β 1 2 β + 2 = 3 2 5 Check solutions, LCD canβ²t be zero 3 Neither make LCD zero, both are solutions 5 Our Solution β β β The LCD can be several factors in these problems. As the LCD gets more complex, it is important to remember the process we are using to solve is still the same. Example 368x + 1)(x + 2) Multiply terms by LCD, (x + 1)(x + 2) x(x + 1)(x + 2) x + 2 + 1(x + 1)(x + 2) x + 1 = 5(x + 1)(x + 2) (x + 1)(x + 2) Reduce fractions 275 x(x + 1) + 1(x + 2) = 5 Distribute x2 + x + x + 2 = 5 Combine like terms x2 + 2x + 2 = 5 Make equatino equal zero Subtract 6 from both sides Factor Set each factor equal to zero Solve each equation 5 5 β β 3 = 0 β 1 x |
2 + 2x (x + 3)(x x + 3 = 0 or x 3 β x = 3 + 2)( ( β 3 or x = 1 Check solutions, LCD canβ²t be zero 3 in (x + 1)(x + 2), it works 2)( 1) = 2 Check β (1 + 1)(1 + 2) = (2)(3) = 6 Check 1 in (x + 1)(x + 2), it works 3 or 1 Our Solution x = β β In the previous example the denominators were factored for us. More often we will need to factor before ο¬nding the LCD Example 369x LCD = (x 11 3x + 2 2) 2) β 1)(x 1)(x β β x2 β β Factor denominator Identify LCD x(x β 1)(x β 2) x β 1 β 1(x β 1)(x β 2) x β 2 = 11(x β 1)(x β 2) (x β 1)(x β 2) Multiply each term by LCD, reduce x(x β x2 2) 1(x 1) = 11 Distribute β β 2x x2 β x + 1 = 11 Combine like terms β 3x + 1 = 11 Make equation equal zero β x2 11 β 3x 11 β 10 = 0 β β (x 5)(x + 2) = 0 5 = 0 or x + 2 = 0 2 2 Check answers, LCD canβ²t be 0 Subtract 11 from both sides Factor Set each factor equal to zero Solve each equation 2 x = 5 or 5 β 2 ( β β 1)( β 2 β 1)(5 2) = (4)(3) = 12 Check 5 in (x 4) = 12 Check β 2 in (x 3)( β 2) = ( β β β 1)(x β 1)(x β 2), it works 2), it works β β β 276 x = 5 or β 2 Our Solution World View Note: Maria Agnesi was the ο¬rst women to publish a math textbook in 1748, it took her over 10 years to write! This textbook covered everything from arithmetic thorugh diο¬erential equations and was over 1,000 pages! If we are subtracting a fraction in the problem, it may be easier to avoid a future sign error by ο¬rst distributing the negative through the numerator. Example 370 Distribute negative through numerator Identify LCD, |
8(x β 3)(x + 2), multiply each term (x β 2)8(x x 3)(x + 2)8(x β x + 2 3)(x + 2) 8(x 5 Β· = 3)(x + 2) β 8 Reduce 8(x β 2)(x + 2) + 8( β x β 2)(x β 3) = 5(x β 3)(x + 2) 8(x2 β 8x2 x2 + x + 6) = 5(x2 4) + 8( 32 β β β 8x2 + 8x + 48 = 5x2 8x + 16 = 5x2 8x 16 0 = 5x2 β β β β β β β FOIL x 6) Distribute β 5x 30 Combine like terms β 5x 30 Make equation equal zero β Subtract 8x and 16 8x 16 β 13x Factor 46 β 0 = (5x 23)(x + 2) Set each factor equal to zero 23 = 0 or x + 2 = 0 Solve each equation 2 2 β β 2 5x = 23 or x = 5 5 β β 5x β + 23 + 23 x = 2 Check solutions, LCD canβ²t be 0 3 8 23 5 β 8( 23 5 3)( + ) = 8( β β 5)(0) = 0 Check Check 23 5 in 8(x β 3)(x + 2), it works 2 in 8(x 3)(x + 2), canβ²t be 0! β β Our Solution 23 5 33 5 or = β 2112 25 x = 23 5 In the previous example, one of the solutions we found made the LCD zero. When this happens we ignore this result and only use the results that make the rational expressions deο¬ned. 277 7.7 Practice - Solving Rational Equations Solve the following equations for the given variable: 1) 3x β x 3) x + 20 β 5 = 5x x β 3 = 2x x β 2 4 β 3 3 4 = 4x + 5 6x 4 β β 3x 1 β 3m + 1 = 3 2 7 5 β x = 12 x 3 x β 7) 9) 2x 3x β 3m 2m β 11) 4 1 β β 7 13 β β 15 = 3x + 8 x2 4 β 17 β 2x + 1 + 2x + 1 1 3 2x = 1 β 8x2 β 4 |
x2 1 β 19) 21) x 23) 25) 27 = 5x + 20 6x + 24 β x x 1 β β 2x x + 1 β 2 x + 1 = 4x2 x2 β 1 3 x + 5 = β 8x2 x2 + 6x + 5 29) x x 31 = β x2 β β 4x2 12x + 27 3 6 + x + 5 x + 3 = β β x2 2x2 3x β 33) 4x + 1 x + 3 + 5x x 3 1 = 8x2 x2 + 2x β β 2) x2 + 6 1 + x x x β 2 1 = 2x β β 6) x x β β 4 1 = 12 3 x + 1 β 8) 6x + 5 2x2 2x β 2 1 β x2 = 3x x2 β 1 4 β 5x 15 = = 3x 3x2 β 4x 2x 7 β 2 β 3 3 10) 12) 14) 16) x + 2 3x β 18) x x β β 20) 3x 5x 5 5 + 5x 7x β β 24) 26) 28 2x x β β 5x2 2 = β x2 + x 6 β 4 = x2 3x 2x β 30) x 3 x + 3 = β 2x2 x2 + 4x + 3 2 3 = x2 x2 + 3x β β 18 β 18 3 β 32 = 9x2 x2 β x 2 2 β β 2x x 3 3 = β β β x2 + 3x 3x2 18 β 34) 3x 1 β x + 6 β 278 2 = 1 x2 + x 1 β 6 β 22 2x + 1 = 2x2 1 3x β 2 β 7.8 Rational Expressions - Dimensional Analysis Objective: Use dimensional analysis to preform single unit, dual unit, square unit, and cubed unit conversions. One application of rational expressions deals with converting units. When we convert units of measure we can do so by multiplying several fractions together in a process known as dimensional analysis. The trick will be to decide what fractions to multiply. When multiplying, if we multiply by 1, the value of the expression does not change. One written as a fraction can look like many diο¬erent things as long as the numerator and denominator are identical in value. Notice the numerator and denominator are |
not identical in appearance, but rather identical in value. Below are several fractions, each equal to one where numerator and denominator are identical in value = 100cm 1m = 1lb 16oz = 1hr 60 min = 60 min 1hr The last few fractions that include units are called conversion factors. We can make a conversion factor out of any two measurements that represent the same distance. For example, 1 mile = 5280 feet. We could then make a conversion 1mi factor 5280ft because both values are the same, the fraction is still equal to one. Similarly we could make a conversion factor 5280ft 1mi. The trick for conversions will 279 be to use the correct fractions. The idea behind dimensional analysis is we will multiply by a fraction in such a way that the units we donβt want will divide out of the problem. We found out when multiplying rational expressions that if a variable appears in the numerator and denominator we can divide it out of the expression. It is the same with units. Consider the following conversion. Example 371. 17.37 miles to feet Write 17.37 miles as a fraction, put it over 1 17.37mi 1??ft??mi 5280ft 1mi 5280ft 1 17.37mi 1 17.37mi 1 17.37 1 To divide out the miles we need miles in the denominator We are converting to feet, so this will go in the numerator Fill in the relationship described above, 1 mile = 5280 feet Divide out the miles and multiply across 91, 713.6ft Our Solution In the previous example, we had to use the conversion factor 5280ft so the miles 1mi would divide out. If we had used 1 mi 5280ft we would not have been able to divide out the miles. This is why when doing dimensional analysis it is very important to use units in the set-up of the problem, so we know how to correctly set up the conversion factor. Example 372. If 1 pound = 16 ounces, how many pounds 435 ounces? 435oz 1 435oz 1?? lbs??oz 435oz 1 1 lbs 16oz Write 435 as a fraction, put it over 1 To divide out oz, put it in the denominator and lbs in numerator Fill in the given relationship, 1 pound = 16 ounces 280 435 1 1 lbs 16 = 435 lbs 16 Divide out oz, multiply across. Divide result 27.1875 lbs Our Solution The same process can be used to convert problems with several units |
in them. Consider the following example. Example 373. A student averaged 45 miles per hour on a trip. What was the studentβs speed in feet per second? 45mi hr 45mi hr 5280ft 1mi 45mi hr 5280ft 1mi 1hr 3600 sec 45 1 5280ft 1 1 3600 sec β²β²perβ²β² is the fraction bar, put hr in denominator To clear mi they must go in denominator and become ft To clear hr they must go in numerator and become sec Divide out mi and hr. Multiply across 237600ft 3600 sec Divide numbers 66 ft per sec Our Solution If the units are two-dimensional (such as square inches - in2) or three-dimensional (such as cubic feet - ft3) we will need to put the same exponent on the conversion factor. So if we are converting square inches (in2) to square ft (ft2), the conversion factor would be squared, the convesion factor. 2 1 ft 12in Example 374.. Similarly if the units are cubed, we will cube Convert 8 cubic feet to yd3 Write 8ft3 as fraction, put it over 1 8ft3 1 To clear ft, put them in denominator, yard in numerator 281 3 3 Because the units are cubed, we cube the conversion factor Evaluate exponent, cubing all numbers and units 8ft3 1??yd??ft 1yd 3ft 8ft3 1 8ft3 1 1yd3 27ft3 Divide out ft3 8 1 1yd3 27 = 8yd3 27 Multiply across and divide 0.296296yd3 Our Solution When calculating area or volume, be sure to use the units and multiply them as well. Example 375. A room is 10 ft by 12 ft. How many square yards are in the room? A = lw = (10ft)(12 ft) = 120ft2 Multiply length by width, also multiply units 120ft2 1 Write area as a fraction, put it over 1 2 120ft2 1??yd??ft 120ft2 1 1yd 3ft Put ft in denominator to clear, square conversion factor Evaluate exponent, squaring all numbers and units 2 120ft2 1 1yd2 9ft2 Divide out ft2 120 1 1yd2 9 = 120yd2 9 Multiply across and divide 13.33yd2 Our solution 282 To focus on the process of conversions, a conversion |
sheet has been included at the end of this lesson which includes several conversion factors for length, volume, mass and time in both English and Metric units. The process of dimensional analysis can be used to convert other types of units as well. If we can identify relationships that represent the same value we can make them into a conversion factor. Example 376. A child is perscribed a dosage of 12 mg of a certain drug and is allowed to reο¬ll his prescription twice. If a there are 60 tablets in a prescription, and each tablet has 4 mg, how many doses are in the 3 prescriptions (original + 2 reο¬lls)? Convert 3 Rx to doses 1 Rx = 60 tab, 1 tab = 4 mg, 1 dose = 12mg Identify what the problem is asking Identify given conversion factors 3Rx 1 3Rx 1 60 tab 1Rx 3Rx 1 60 tab 1Rx 4mg 1 tab 3Rx 1 60 tab 1Rx 4mg 1 tab 1 dose 12mg 3 1 60 1 4 1 1 dose 12 Write 3Rx as fraction, put over 1 Convert Rx to tab, put Rx in denominator Convert tab to mg, put tab in denominator Convert mg to dose, put mg in denominator Divide out Rx, tab, and mg, multiply across 720 dose 12 Divide 60 doses Our Solution World View Note: Only three countries in the world still use the English system commercially: Liberia (Western Africa), Myanmar (between India and Vietnam), and the USA. 283 Conversion Factors Length English 12 in = 1 ft 1 yd = 3 ft 1 yd = 36 in 1 mi = 5280 ft Metric 1000 mm = 1 m 10 mm = 1 cm 100 cm = 1 m 10 dm = 1 m 1 dam = 10 m 1 hm = 100 m 1 km = 1000 m English/Metric 2.54 cm = 1 in 1 m = 3.28 ft 1.61 km = 1 mi Volume English 1 c = 8 oz 1 pt = 2 c 1 qt = 2 pt 1 gal = 4 qt Metric 1 mL = 1 cc = 1 cm3 1 L = 1000 mL 1 L = 100 cL 1 L = 10 dL 1000 L = 1 kL English/Metric 16.39 mL = 1 in3 1.06 qt = 1 L 3.79 L = 1gal Area English 1 ft2 = 144 in2 1 yd2 = 9 ft2 1 acre = 43, |
560 ft2 640 acres = 1 mi2 Metric 1 a = 100 m2 1 ha = 100 a English/Metric 1 ha = 2.47 acres English 1 lb = 16 oz 1 T = 2000 lb Weight (Mass) Metric 1 g = 1000 mg 1 g = 100 cg 1000 g = 1 kg 1000 kg = 1 t English/Metric 28.3 g = 1 oz 2.2 lb = 1 kg Time 60 sec = 1 min 60 min = 1 hr 3600 sec = 1 hr 24 hr = 1 day 284 7.8 Practice - Dimensional Analysis Use dimensional analysis to convert the following: 1) 7 mi. to yards 2) 234 oz. to tons 3) 11.2 mg to grams 4) 1.35 km to centimeters 5) 9,800,000 mm (milimeters) to miles 6) 4.5 ft2 to square yards 7) 435,000 m2 to sqaure kilometers 8) 8 km2 to square feet 9) 0.0065 km3 to cubic meters 10) 14.62 in3 to cubic centimeters 11) 5,500 cm3 to cubic yards 12) 3.5 mph (miles per hour) to feet per second 13) 185 yd. per min. to miles per hour 14) 153 ft/s (feet per second) to miles per hour 15) 248 mph to meters per second 16) 186,000 mph to kilometers per year 17) 7.50 T/yd2 (tons per square yard) to pounds per square inch 18) 16 ft/s2 to kilometers per hour squared 285 Use dimensional analysis to solve the following: 19) On a recent trip, Jan traveled 260 miles using 8 gallons of gas. How many miles per 1-gallon did she travel? How many yards per 1-ounce? 20) A chair lift at the Divide ski resort in Cold Springs, WY is 4806 feet long and takes 9 minutes. What is the average speed in miles per hour? How many feet per second does the lift travel? 21) A certain laser printer can print 12 pages per minute. Determine this printerβs output in pages per day, and reams per month. (1 ream = 5000 pages) 22) An average human heart beats 60 times per minute. If an average person lives to the age of 75, how many times does the average heart beat in a lifetime? 23) Blood sugar levels are measured in miligrams of gluclose per dec |
iliter of blood volume. If a personβs blood sugar level measured 128 mg/dL, how much is this in grams per liter? 24) You are buying carpet to cover a room that measures 38 ft by 40 ft. The carpet cost S18 per square yard. How much will the carpet cost? 25) A car travels 14 miles in 15 minutes. How fast is it going in miles per hour? in meters per second? 26) A cargo container is 50 ft long, 10 ft wide, and 8 ft tall. Find its volume in cubic yards and cubic meters. 27) A local zoning ordinance says that a houseβs βfootprintβ (area of its ground ο¬oor) cannot occupy more than 1 1 3 acre lot, what is the maximum allowed footprint for your house in square feet? in square inches? (1 acre = 43560 ft2) 4 of the lot it is built on. Suppose you own a 28) Computer memory is measured in units of bytes, where one byte is enough memory to store one character (a letter in the alphabet or a number). How many typical pages of text can be stored on a 700-megabyte compact disc? Assume that one typical page of text contains 2000 characters. (1 megabyte = 1,000,000 bytes) 29) In April 1996, the Department of the Interior released a βspike ο¬oodβ from the Glen Canyon Dam on the Colorado River. Its purpose was to restore the river and the habitants along its bank. The release from the dam lasted a week at a rate of 25,800 cubic feet of water per second. About how much water was released during the 1-week ο¬ood? 30) The largest single rough diamond ever found, the Cullinan diamond, weighed 3106 carats; how much does the diamond weigh in miligrams? in pounds? (1 carat - 0.2 grams) 286 Chapter 8 : Radicals 8.1 Square Roots..........................................................................................288 8.2 Higher Roots..........................................................................................292 8.3 Adding Radicals.....................................................................................295 8.4 Multiply and Divide Radicals................................................................298 8.5 Rationalize Denominators......................................................................303 8.6 Rational Exponents................................................................................310 8.7 Radicals of Mixed Index.........................................................................314 8.8 Complex Numbers..................................................................................318 287 8.1 Radicals - Square Roots Objective: Simplify expressions with square |
roots. Square roots are the most common type of radical used. A square root βunsquaresβ a number. For example, because 52 = 25 we say the square root of 25 is 5. The square root of 25 is written as 25β. World View Note: The radical sign, when ο¬rst used was an R with a line through the tail, similar to our perscription symbol today. The R came from the latin, βradixβ, which can be translated as βsourceβ or βfoundationβ. It wasnβt until the 1500s that our current symbol was ο¬rst used in Germany (but even then it was just a check mark with no bar over the numbers! The following example gives several square roots: Example 377. 1β = 1 4β = 2 9β = 3 121β 625β = 11 = 25 = Undeο¬ned 81β β 81β β The ο¬nal example, is currently undeο¬ned as negatives have no square root. This is because if we square a positive or a negative, the answer will be positive. Thus we can only take square roots of positive numbers. In another lesson we will deο¬ne a method we can use to work with and evaluate negative square roots, but for now we will simply say they are undeο¬ned. 8β on Not all numbers have a nice even square root. For example, if we found our calculator, the answer would be 2.828427124746190097603377448419... and even this number is a rounded approximation of the square root. To be as accurate as possible, we will never use the calculator to ο¬nd decimal approximations of square roots. Instead we will express roots in simplest radical form. We will do this using a property known as the product rule of radicals Product Rule of Square Roots: a bβ Β· = aβ bβ Β· We can use the product rule to simplify an expression such as it into two roots, by spliting 5β, and simplifying the ο¬rst root, 6 5β. The trick in this 36β Β· 5β 36 Β· 288 process is being able to translate a problem like. There are several ways this can be done. The most common and, with a bit of practice, the |
fastest method, is to ο¬nd perfect squares that divide evenly into the radicand, or number under the radical. This is shown in the next example. into Β· 5β 36 180β Example 378. 75β 3β 25 Β· 3β Β· 5 3β 25β 75 is divisible by 25, a perfect square Split into factors Product rule, take the square root of 25 Our Solution If there is a coeο¬cient in front of the radical to begin with, the problem merely becomes a big multiplication problem. Example 379. 5 63β 7β 5 9 Β· 7β Β· 3 7β Β· 15 7β 5 9β 5 63 is divisible by 9, a perfect square Split into factors Product rule, take the square root of 9 Multiply coeο¬cients Our Solution As we simplify radicals using this method it is important to be sure our ο¬nal answer can be simpliο¬ed no more. Example 380. 9β 72β 8β 9 Β· 8β Β· 3 8β 2β 3 4 Β· 2β Β· 2 2β 3 4β 3 Β· 72 is divisible by 9, a perfect square Split into factors Product rule, take the square root of 9 But 8 is also divisible by a perfect square, 4 Split into factors Product rule, take the square root of 4 Multiply 289 6 2β Our Solution. The previous example could have been done in fewer steps if we had noticed that 72 = 36 2, but often the time it takes to discover the larger perfect square is more than it would take to simplify in several steps. Β· Variables often are part of the radicand as well. When taking the square roots of x8β = x4, because we variables, we can divide the exponent by 2. For example, divide the exponent of 8 by 2. This follows from the power of a power rule of expoennts, (x4)2 = x8. When squaring, we multiply the exponent by two, so when taking a square root we divide the exponent by 2. This is shown in the following example. Example 381. 5 9β 2β Β· Β· β Β· 5 9 p 18x4y6z10 β 2x4y6z10 5 β x4β z10β y6 p Β· Β· 3 |
x2y3z5 2β 5 p Β· 15x2y3z5 2β β β 18 is divisible by 9, a perfect square Split into factors Product rule, simplify roots, divide exponents by 2 Multiply coeο¬cients Our Solution We canβt always evenly divide the exponent on a variable by 2. Sometimes we have a remainder. If there is a remainder, this means the remainder is left inside the radical, and the whole number part is how many are outside the radical. This is shown in the following example. Example 382. 4 p 20x5y9z6 5x5y9z6 z6β y9 p Β· 2x2y4z3 5xyβ Β· Β· p x5β 4β 5β Β· Β· 20 is divisible by 4, a perfect square Split into factors Simplify, divide exponents by 2, remainder is left inside Our Solution 290 8.1 Practice - Square Roots Simplify. 1) 245β 3) 36β 5) 12β 7) 3 12β 9) 6 128β 11) 13) 8 392β β 192nβ 15) β 196v2 17) β 252x2 19) 21) 23) β 100k4 7 64x4β 5 36mβ β β β 25) 45x2y2 p p 27) 29) 16x3y3 320x4y4 p 31) 6 80xy2 p 33) 5 245x2y3 35) 37) p β β 2 180u3v 8 180x4y2z4 β 39) 2 p 80hj4k 41) p 4 β p 54mnp2 2) 125β 4) 196β 6) 72β 8) 5 32β 10) 7 128β 12) 14) 7 63β β 343bβ 16) β 100n3 18) β 200a3 20) 22) β β 4 175p4 p 2 128nβ 24) 8 112p2 p β 72a3b4 26) 28) β 512a4b2 30) β 512m4n3 β 32) 8 98mn 34) 2 72x2y2 72x3y4 36) p 5 β β 38) 6 50a4bc2 p 40 |
) 42) β β p 8 32xy2z3 32m2p4q p 291 8.2 Radicals - Higher Roots Objective: Simplify radicals with an index greater than two. While square roots are the most common type of radical we work with, we can take higher roots of numbers as well: cube roots, fourth roots, ο¬fth roots, etc. Following is a deο¬nition of radicals. amβ = b if bm = a The small letter m inside the radical is called the index. It tells us which root we are taking, or which power we are βun-doingβ. For square roots the index is 2. As this is the most common root, the two is not usually written. World View Note: The word for root comes from the French mathematician Franciscus Vieta in the late 16th century. The following example includes several higher roots. Example 383. 3β = 5 125 814β = 3 325β = 2 3β 7β 4 2 = = 64 β 128 β = undeο¬ned β β 16 4β β We must be careful of a few things as we work with higher roots. First its impor814β = 3. This is tant not to forget to check the index on the root. because 92 = 81 and 34 = 81. Another thing to watch out for is negatives under roots. We can take an odd root of a negative number, because a negative number raised to an odd power is still negative. However, we cannot take an even root of a negative number, this we will say is undeο¬ned. In a later section we will discuss how to work with roots of negative, but for now we will simply say they are undeο¬ned. 81β = 9 but We can simplify higher roots in much the same way we simpliο¬ed square roots, using the product property of radicals. Product Property of Radicals: abmβ = amβ bmβ Β· Often we are not as familiar with higher powers as we are with squares. It is important to remember what index we are working with as we try and work our way to the solution. 292 Example 384. We are working with a cubed root, want third powers 543β 23 = 8 Test 2, 23 = 8, 54 is not divisible by 8. 33 = 27 Test 3, 33 = |
27, 54 is divisible by 27! 3β 273β Β· 3 Write as factors Product rule, take cubed root of 27 Our Solution 2 Β· 23β 23β 27 Just as with square roots, if we have a coeο¬cient, we multiply the new coeο¬cients together. Example 385. We are working with a fourth root, want fourth powers 484β 3 24 = 16 Test 2, 24 = 16, 48 is divisible by 16! 4β 16 Write as factors Product rule, take fourth root of 16 Multiply coeο¬cients Our Solution 3 4β 3 3 Β· 34β 34β 34β 16 Β· 2 3 6 Β· We can also take higher roots of variables. As we do, we will divide the exponent on the variable by the index. Any whole answer is how many of that varible will come out of the square root. Any remainder is how many are left behind inside the square root. This is shown in the following examples. Example 386. 5 x25y17z3 y2z3 5 x5y3 p Divide each exponent by 5, whole number outside, remainder inside Our Solution p In the previous example, for the x, we divided 25 remain inside. For the y, we divided 17 inside. For the z, when we divided 3 following example includes integers in our problem. 5 = 5 R 0, so x5 came out, no xβs 5 = 3 R 2, so y3 came out, and y2 remains 5 = 0R 3, all three or z3 remained inside. The Example 387. 3β 2 3β 8 2 2ab2 4ab2 40a4b8 5a4b8 5ab2 5ab2 Β· 3β 3β 2 Β· Looking for cubes that divide into 40. The number 8 works! Take cube root of 8, dividing exponents on variables by 3 Remainders are left in radical. Multiply coeο¬cients Our Solution 293 8.2 Practice - Higher Roots Simplify. 1) 3β 625 3) 3β 750 5) 3β 875 2) 3β 375 4) 3β 250 6) 243β 4 964β 7) β 4β 9) 6 112 11) 13) β 4β 4β 112 648a2 15) 5β |
224n3 5 17) 224p5 p β β β 3 p 3 p 3 19) 21) 23) 25) 27) 29) 7β 3 896r 3β 2 48v7 β 3β 7 320n6 135x5y3 32x4y4 β β 256x4y6 p 3 31) 7 81x3y7 β p 3β 33) 2 375u2v8 8 484β 8) β 10) 3 484β 4β 12) 5 243 14) 4β 64n3 16) 5β 18) 6β 96x3 β 256x6 20) 7β 8 384b8 β 3β 22) 4 250a6 24) 26) β 3β 3β 512n6 64u5v3 28) 3β 1000a4b5 3 30) 189x3y6 32) 3 56x2y8 4 p β 3β 34) 8 750xy p β 135xy3 35) 3β 3 192ab2 3 36) 3 β 3 37) 6 39) 6 41) 7 p 4 p 4 p 54m8n3p7 β 648x5y7z2 128h6j8k8 38) 40) 42) 4 p 6 β 80m4p7q4 p 4β 6 405a5b8c 4 6 324x7yz7 β β p 294 8.3 Radicals - Adding Radicals Objective: Add like radicals by ο¬rst simplifying each radical. Adding and subtracting radicals is very similar to adding and subtracting with variables. Consider the following example. Example 388. 5x + 3x β 2x Combine like terms 6x Our Solution 5 11β + 3 11β β 2 11β 6 11β Combine like terms Our Solution 11β it was just like combining Notice that when we combined the terms with terms with x. When adding and subtracting with radicals we can combine like radicals just as like terms. We add and subtract the coeο¬cients in front of the 295 radical, and the radical stays the same. This is shown in the following example. Example 389. 7 65β + 4 35β β 8 9 65β 35β + 65β 35β 5 β Combine like radicals 7 Our Solution 65β + 65β and |
4 35β 8 35β β We cannot simplify this expression any more as the radicals do not match. Often problems we solve have no like radicals, however, if we simplify the radicals ο¬rst we may ο¬nd we do in fact have like radicals. Example 390. 5 45β + 6 18β 5β 2β 3 5β + 6 5 β Β· Β· 15 5β + 18 2β β 2β 2 49 2 2 98β + 20β 5β + 4 Β· Β· 7 2β + 2 5β Β· 14 2β + 2 5β 17 5β + 4 2β β Simplify radicals, ο¬nd perfect square factors Take roots where possible Multiply coeο¬cients Combine like terms Our Solution World View Note: The Arab writers of the 16th century used the symbol similar to the greater than symbol with a dot underneath for radicals. This exact process can be used to add and subtract radicals with higher indices Example 391. 4 27 3 12 543β 2 Β· β 23β β 23β β 9 9 3β 4 4 Β· 163β + 5 9 3β 8 + 5 2 Β· 23β + 5 2 Β· 23β + 5 18 23β + 5 6 β β 93β 93β 93β 93β 93β Simplify each radical, ο¬nding perfect cube factors Take roots where possible Multiply coeο¬cients Combine like terms 12 Our Solution 23β 23β 18 β 296 8.3 Practice - Adding Radicals Simiplify 1) 2 5β + 2 5β + 2 5β 3 2β + 3 5β + 3 5β 3) β 5) 2 6β 2 6β 6β β β 7) 3 6β + 3 5β + 2 5β β 2) 4) 6) 8) β β β β 2 3β β 3 6β 3 6β 3 3β β 2 6β 3β β β 3 3β + 2 3β 2 3β β 2 3β β 5β + 2 3β 9) 2 2β 3 18β 2β β 10) 54β β 3 6β + 3 27β β |
5β 12) 5β β β β 14) 2 20β + 2 20β 2 54β 3β β 3 27β + 2 3β 12β β 16) 18) 20) 22) β β β β 2 2β β 3 18β 3 8β β 5β β 2β + 3 8β + 3 6β 8β + 2 8β + 2 8β 3 6β + 2 18β β 24) 2 6β 54β 3 27β 3β β β β 3β 26) 3 135 3β 81 β β 3 44β + 3 324 4β 3β 135 + 2 644β 30) 2 64β + 2 44β + 3 64β 4β 2 243 964β + 2 964β 32) β 34) 2 484β β 4β 3 405 3 484β β β 4β 162 7β 3 768 7β + 2 384 + 3 57β β 3 37β β 7β 2 256 7β 2 256 3 27β β β 7β 640 β β 3 6β 11) β 12β + 3 3β β 13) 3 2β + 2 8β 3 18β β 15) 3 18β 2β 3 2β β 3 6β β 2 18β β 3 6β β 3 8β β β 17) 19) β β 3β + 3 6β 20β + 2 20β 2 24β 21) β 23) 3 24β β 2 6β + 2 6β + 2 20β β 3 27β + 2 6β + 2 8β 2 163β + 2 163β + 2 23β 25) β 4β 27) 2 243 29) 3 24β β β 2 24β 4β 243 β 4β + 3 324 31) 4β 324 β β 33) 2 24β + 2 34β + 3 644β 3 44β 34β β 4β 2 243 34β β 28) β 3 65β 35) β 5β 37) 2 160 β β 645β + 2 192 5 |
β 5β 2 64 β 5β 2 192 5β 160 β 39) 6β 256 2 46β β β β 6β 3 320 160 5β β β 6β 2 128 β 36) 38) β β 297 8.4 Radicals - Multiply and Divide Radicals Objective: Multiply and divide radicals using the product and quotient rules of radicals. Multiplying radicals is very simple if the index on all the radicals match. The prodcut rule of radicals which we have already been using can be generalized as follows: Product Rule of Radicals: a bmβ Β· c dmβ = ac bdmβ Another way of stating this rule is we are allowed to multiply the factors outside the radical and we are allowed to multiply the factors inside the radicals, as long as the index matches. This is shown in the following example. Example 392. β 5 14β 4 6β Β· 20 84β 21β Β· 2 21β Β· 40 21β β 20 4 20 β β β Multiply outside and inside the radical Simplify the radical, divisible by 4 Take the square root where possible Multiply coeο¬cients Our Solution The same process works with higher roots Example 393. 3β 2 18 Β· 12 3β 3β 6 3β 27 3 Β· 36 Β· 3β 3β 15 270 10 10 10 12 12 Multiply outside and inside the radical Simplify the radical, divisible by 27 Take cube root where possible Multiply coeο¬cients Our Solution When multiplying with radicals we can still use the distributive property or FOIL just as we could with variables. Example 394. 7 6β (3 10β 21 60β 21 4 21 15β β Β· 2 15β β 42 15β Β· β β 35 9 35 5 15β ) Distribute, following rules for multiplying radicals 35 90β Simplify each radical, ο¬nding perfect square factors 10β Take square root where possible Β· 3 10β Multiply coeο¬cients Β· 105 10β Our Solution β Example 395. ( 5β β 2 3β )(4 10β + 6 6β ) FOIL, following rules for multiplying radicals 298 4 50β + 6 30β 8 30β Β· 2β 4 25 |
4 + 6 30β 5 2β + 6 30β β β 20 2β + 6 30β Β· β 8 30β 8 30β β β 8 30β 16 2β β β β 12 9 12 12 18β 2β Β· 3 2β Β· 36 2β 2 30β β β Simplify radicals, ο¬nd perfect square factors Take square root where possible Multiply coeο¬cients Combine like terms Our Solution World View Note: Clay tablets have been discovered revealing much about Babylonian mathematics dating back from 1800 to 1600 BC. In one of the tables there is an approximation of 2β accurate to ο¬ve decimal places (1.41421) Example 396. (2 5β β 16 35β 14 10β 14 10β 14 10β β β 14 10β β 16 35β 16 35β β β 16 35β 3 6β )(7 2β 21 12β 3β Β· 2 3β Β· 42 3β β 21 4 21 8 7β ) β 24 42β 24 42β 24 42β 24 42β β β β β β β FOIL, following rules for multiplying radicals Simplify radicals, ο¬nd perfect square factors Take square root where possible Multiply coeο¬cient Our Solution As we are multiplying we always look at our ο¬nal solution to check if all the radicals are simpliο¬ed and all like radicals or like terms have been combined. Division with radicals is very similar to multiplication, if we think about division as reducing fractions, we can reduce the coeο¬cients outside the radicals and reduce the values inside the radicals to get our ο¬nal solution. Quotient Rule of Radicals: a c bmβ dmβ = a c b d m r Example 397. 15 20 3β 108 23β Reduce 15 20 and 3β 108 2β by dividing by 5 and 2 respectively 3 543β 4 3β 3 2 27 4 Β· 3 Β· 23β 3 4 9 23β 4 Simplify radical, 54 is divisible by 27 Take the cube root of 27 Multiply coeο¬cients Our Solution There is one catch to dividing with radicals, it is considered bad practice to have a radical in |
the denominator we will rationalize it, or clear out any radicals in the denominator. in the denominator of our ο¬nal answer. If there is a radical 299 We do this by multiplying the numerator and denominator by the same thing. The problems we will consider here will all have a monomial in the denominator. The way we clear a monomial radical in the denominator is to focus on the index. The index tells us how many of each factor we will need to clear the radical. For example, if the index is 4, we will need 4 of each factor to clear the radical. This is shown in the following examples. Example 398. 6β 5β 6β 5β 5β 5β! Index is 2, we need two ο¬ves in denominator, need 1 more Multiply numerator and denominator by 5β 30β 5 Our Solution Example 399. 3 4β 11 24β 3 114β 24β 4β 23 4β! 23 Index is 4, we need four twos in denominator, need 3 more Multiply numerator and denominator by 23 4β 88 3 4β 2 Our Solution Example 400. 4 7 23β 253β 4 7 23β 3β 52 4 7 23β 3β 52 53β 53β! 4 4 103β 5 7 Β· 103β 35 The 25 can be written as 52. This will help us keep the numbers small Index is 3, we need three ο¬ves in denominator, need 1 more Multiply numerator and denominator by 53β Multiply out denominator Our Solution 300 The previous example could have been solved by multiplying numerator and 3β. However, this would have made the numbers very large denominator by and we would have needed to reduce our soultion at the end. This is why re53β was the better way to simwriting the radical as plify. 3β and multiplying by just 252 52 We will also always want to reduce our fractions (inside and out of the radical) before we rationalize. Example 401. 6 14β 12 22β 7β 2 11β 7β 2 11β 11β 11β! Reduce coeο¬cients and inside radical Index is 2, need two elevens, need 1 more Mult |
iply numerator and denominator by 11β 77β 11 2 Β· 77β 22 Multiply denominator Our Solution The same process can be used to rationalize fractions with variables. Example 402. Reduce coeο¬cients and inside radical 6x3y4z 10xy6z3 4 18 4 p8 p 9 4 4 4β 3x2 5y2z3 Index is 4. We need four of everything to rationalize, three more ο¬ves, two more y β²s and one more z. 9 4 4β 3x2 5y2z3 4 p 4 9! 4 p 53y2z 4 p53y2z p 375x2y2z 5yz p4 Β· Multiply numerator and denominator by 4 53y2z p Multiply denominator 4 9 375x2y2z p20yz Our Solution 301 8.4 Practice - Multiply and Divide Radicals Multiply or Divide and Simplify. 1) 3 5β 4 16β Β· β 3) 12mβ 15mβ Β· 3β 3β 4x3 2x4 Β· 6β ( 2β + 2) 5) 7) 9) 5 15β (3 3β + 2) β 11) 5 10β (5n + 2β ) 13) (2 + 2 2β )( 3 + 2β ) 15) ( 5β β β 5)(2 5β 1) β 17) ( 2aβ + 2 3aβ )(3 2aβ + 5aβ ) 19) ( 5 4 3β )( 3 β β 4 3β ) β β 12β 5 100β 21) 23) 5β 4 125β 25) 10β 6β 27) 2 4β 3 3β 29) 5x2 3x3y3 4 31) p 2p2 p 3pβ 33) 3 103β 5 273β 35) 3β 5 3β 4 4 4β 37) 5 5r4 4β 8r2 5 10β 2) 4) β 5r3β Β· β 15β Β· 5 10r2β 3β 6) 3 4a4 3β 10a3 Β· 10β |
( 5β + 2β ) 8) 10) 5 15β (3 3β + 2) 12) 15β ( 5β 3 3vβ ) β 14) ( β 2 + 3β )( β 5 + 2 3β ) 16) (2 3β + 5β )(5 3β + 2 4β ) 18) ( β 2 2pβ + 5 5β )( 5pβ + 5pβ ) 20) (5 2β 1)( β β 2mβ + 5) 22) 24) 26) 15β 2 4β 12β 3β 2β 3 5β 28) 4 3β 15β 30) 5 4 3xy4 p 8n2β 10nβ 32) 34) 153β 643β 36) 24β 2 644β 38) 4 64m4n2 4β 302 8.5 Radicals - Rationalize Denominators Objective: Rationalize the denominators of radical expressions. It is considered bad practice to have a radical in the denominator of a fraction. When this happens we multiply the numerator and denominator by the same thing in order to clear the radical. In the lesson on dividing radicals we talked about how this was done with monomials. Here we will look at how this is done with binomials. If the binomial is in the numerator the process to rationalize the denominator is essentially the same as with monomials. The only diο¬erence is we will have to distribute in the numerator. Example 403. 9 3β β 2 6β ( 3β 9) β 2 6β 6β 6β! Want to clear 6β in denominator, multiply by 6β We will distribute the 6β through the numerator 303 18β 9 6β β 6 2 Β· 9 6β 2β 9 Β· β 12 3 2β 2β 9 6β 3 6β β 12 β 4 Simplify radicals in numerator, multiply out denominator Take square root where possible Reduce by dividing each term by 3 Our Solution It is important to remember that when reducing the fraction we cannot reduce with just the 3 and 12 or just the 9 and 12. When we have addition or subtraction in the numerator or |
denominator we must divide all terms by the same number. The problem can often be made easier if we ο¬rst simplify any radicals in the problem. 2 20x5β 12x2β β 18xβ Simplify radicals by ο¬nding perfect squares β 2 4 3x2 4 Β· Β· β 5x3 β 2xβ 9 Β· 2 Β· 2x2 5xβ β 3 2xβ 2x 3β Simplify roots, divide exponents by 2. Multiply coeο¬cients 4x2 5xβ 2x 3β β 3 2xβ Multiplying numerator and denominator by 2xβ (4x2 5xβ 2x 3β ) β 3 2xβ 4x2 10x2β 3 Β· β 2x 2xβ 2xβ! 2x 6xβ Distribute through numerator Simplify roots in numerator, multiply coeο¬cients in denominator 4x3 10β β 6x 2x 6xβ Reduce, dividing each term by 2x 304 2x2 10β 3x β 6xβ Our Solution As we are rationalizing it will always be important to constantly check our problem to see if it can be simpliο¬ed more. We ask ourselves, can the fraction be reduced? Can the radicals be simpliο¬ed? These steps may happen several times on our way to the solution. 5 β If the binomial occurs in the denominator we will have to use a diο¬erent strategy 2, if we were to multiply the denominator by to clear the radical. Consider 3β 5 3β. We have 3β we would have to distribute it and we would end up with 3 not cleared the radical, only moved it to another part of the denominator. So our current method will not work. Instead we will use what is called a conjugate. A conjugate is made up of the same terms, with the opposite sign in the middle. 3β + So for our example with 5. The advantage of a conjugate is when we multiply them together we have 5)( 3β + 5), which is a sum and a diο¬erence. We know when we multiply ( 3β 3β and 5, with subtraction in |
the these we get a diο¬erence of squares. Squaring middle gives the product 3 22. Our answer when multiplying conjugates 25 = will no longer have a square root. This is exactly what we want. 5 in the denominator, the conjugate would be 3β β β β β β Example 404. 2 3β β 5 2 3β β 3β + 5 3β + 5! 5 Multiply numerator and denominator by conjugate Distribute numerator, diο¬erence of squares in denominator 2 3β + 10 25 3 β 2 3β + 10 22 β β 5 3β 11 β Simplify denoinator Reduce by dividing all terms by 2 β Our Solution In the previous example, we could have reduced by dividng by 2, giving the solution 3β + 5 11, both answers are correct. β Example 405. 15β 5β + 3β Multiply by conjugate, 5β 3β β 305 15β 5β + 3β 5β 5β 3β 3β! β β 75β 5 45β 3 β β 3β 25 Β· 5β 9 Β· β 2 Distribute numerator, denominator is diο¬erence of squares Simplify radicals in numerator, subtract in denominator Take square roots where possible 5 3β 3 5β β 2 Our Solution Example 406. 2 3xβ 5x3β 4 β 2 3xβ 5x3β 4 β 4 + 5x3β 4 + 5x3β! 8 3xβ + 2 15x4β 5x3 16 β 8 3xβ + 2x2 15β 5x3 16 β Multiply by conjugate, 4 + 5x3β Distribute numerator, denominator is diο¬erence of squares Simplify radicals where possible Our Solution The same process can be used when there is a binomial in the numerator and denominator. We just need to remember to FOIL out the numerator. Example 407. 5β 3β 3 2 β β 5β 3β 2 + 3β 2 + 3β! 3 2 β β Multiply by conjugate, 2 + 3β FOIL in numerator, denominator is di |
ο¬erence of squares 6 + 3 3β 2 5β 3 β 4 β 15β β Simplify denominator 6 + 3 3β 2 5β 1 β β 15β Divide each term by 1 6 + 3 3β 2 5β 15β β β Our Solution 306 Example 408. 2 5β 3 7β β 5 6β + 4 2β Multiply by the conjugate, 5 6β 4 2β β 3 7β 2 5β β 5 6β + 4 2β 5 6β 5 6β 4 2β 4 2β! β β FOIL numerator, denominator is diο¬erence of squares 10 30β 10 30β 10 30β β β β 8 10β β 6 25 β Β· 15 42β + 12 14β 16 2 Β· 8 10β β 150 β 15 42β + 12 14β 32 8 10β 15 42β + 12 14β β 118 Multiply in denominator Subtract in denominator Our Solution The same process is used when we have variables Example 409. 3x 2xβ + 4x3β 3xβ 5x β Multiply by the conjugate, 5x + 3xβ 3x 2xβ + 4x3β 3xβ 5x β 5x + 3xβ 5x + 3xβ! FOIL in numerator, denominator is diο¬erence of squares 15x2 2xβ + 3x 6x2β + 5x 4x3β + 12x4β 25x2 3x β Simplify radicals 15x2 2xβ + 3x2 6β + 10x2 xβ + 2x2 3β 25x2 3x β Divide each term by x 15x 2xβ + 3x 6β + 10x xβ + 2x 3β 25x 3 β Our Solution World View Note: During the 5th century BC in India, Aryabhata published a treatise on astronomy. His work included a method for ο¬nding the square root of numbers that have many digits. 307 8.5 Practice - Rationalize Denominators Simplify. 1) 4 + 2 3β |
9β 3) 4 + 2 3β 5 4β 5) 2 5 5β β 4 13β 7) 2β 3 3β β 3β 9) 5 3 5β + 2β 11) 2 5 + 2β 13) 3 3 3β 4 β 15) 4 3 + 5β 17) 19) 21) 23) 4 4 2β β 4 β 1 1 + 2β 14β 7β 2 β 2β β abβ bβ a β aβ β 25) a + abβ aβ + bβ 27) 2 + 6β 2 + 3β 29) a bβ β a + bβ 31) 33) 35) 3 2β a a bβ 6 β β β 2 3β b b aβ 5β 2 β 3 + 5β β 2) β 4 + 3β 4 9β 4) 2 3β β 2 16β 2 6) 5β + 4 4 17β 8) 5β 2β β 3 6β 10) 5 3β + 4 5β 12) 14) 5 β 2β 2 3β 4 2β β 2 16) 2 2 5β + 2 3β 18) 4 β 5β 4 3β 20) 3 + 3β 3β 1 β 22) 2 + 10β 2β + 5β 24) 7β 14β β 14β + 7β 26) a + abβ aβ + bβ 28) 2 5β + 3β 3β 1 β 30) 32) a b β aβ + bβ ab β b aβ a bβ 34) 4 2β + 3 3 2β + 3β 36) β 1 + 5β 2 5β + 5 2β 308 37) 5 2β + 3β 5 + 5 2β 38) 3β + 2β 2β 2 3β β 309 8.6 Radicals - Rational Exponents Objective: Convert between radical notation and exponential notation and simplify expressions with rational exponents using the properties of exponents. When we simplify radicals with exponents, we divide the exponent by the |
index. Another way to write division is with a fraction bar. This idea is how we will deο¬ne rational exponents. Deο¬nition of Rational Exponents: a n m = ( amβ )n The denominator of a rational exponent becomes the index on our radical, likewise the index on the radical becomes the denominator of the exponent. We can use this property to change any radical expression into an exponential expression. Example 410. ( x5β )3 = x 3 5 1 ( a7β )3 3 7 = aβ 5 6 ( 3x6β )5 = (3x) 1 3β )2 = (xy)β ( xy 2 3 Index is denominator Negative exponents from reciprocals We can also change any rational exponent into a radical expression by using the denominator as the index. Example 411. 5 3 = ( a3β )5 a (2mn) 2 7β 7 = ( 2mn )2 4 xβ 5 = 1 ( x5β )4 2 (xy)β 9 = 1 ( xy 9β )2 Index is denominator Negative exponent means reciprocals World View Note: Nicole Oresme, a Mathematician born in Normandy was the ο¬rst to use rational exponents. He used the notation 1 3β’ ever his notation went largely unnoticed. 9p to represent 9 3. How- 1 The ability to change between exponential expressions and radical expressions allows us to evaluate problems we had no means of evaluating before by changing to a radical. Example 412. 4 27β 3 Change to radical, denominator is index, negative means reciprocal 1 3β )4 ( 27 Evaluate radical 310 1 (3)4 1 81 Evaluate exponent Our solution The largest advantage of being able to change a radical expression into an exponential expression is we are now allowed to use all our exponent properties to simplify. The following table reviews all of our exponent properties. Properties of Exponents aman = am+n (ab)m = ambm am an = am n β (am)n = amn a b m = am bm a0 = 1 aβ m = 1 am 1 m = am aβ a b m β = bm am When adding and subtracting with fractions we need to be sure to have a common denominator. When multiplying we only need to multiply the numerators together and denominators together. The following examples show several |
diο¬erent problems, using diο¬erent properties to simplify the rational exponents. Example 413 Need common denominator on aβ²s(6) and bβ²s(10) 4 5 1 2 a 6 b 10 a 6 b 10 Add exponents on aβ²s and bβ²s 5 7 a 6 b 10 Our Solution Example 414. Example 415 Multiply 3 4 by each exponent x 4 y 10 Our Solution x2y 2 3 2x 7 Β· 2 y0 x 1 2 y 5 6 In numerator, need common denominator to add exponents 311 4 x 2 y 4 6 x 2x 7 Β· 2 y0 1 2 y 5 6 Add exponents in numerator, in denominator, y0 = 1 5 2x 2 y 9 6 7 2 x Subtract exponents on x, reduce exponent on y 2xβ 1y 3 2 Negative exponent moves down to denominator 3 2 2y x Our Solution Example 416. 1 25x 3 y 2 5 1 2 β 4 9x 5 yβ 3 2 ο£Ά Using order of operations, simplify inside parenthesis ο¬rst Need common denominators before we can subtract exponents  ο£ ο£Έ 1 2 β 4 10 15 10 ο£Ά ο£Έ Subtract exponents, be careful of the negative: 15 4 10 10 β 19 10 4 10 15 10 β + = = 5 25x 15 y 12 15 yβ 9x  ο£ 19 10 7 15 y 25xβ 9 9 7 25xβ 15 y 1 2 β! 1 2 19 10! 1 2 9 1 7 25 2 xβ 30 y 3x 5y 19 20 7 30 19 20 The negative exponent will ο¬ip the fraction The exponent 1 2 goes on each factor 1 1 Evaluate 9 2 and 25 2 and move negative exponent Our Solution It is important to remember that as we simplify with rational exponents we are using the exact same properties we used when simplifying integer exponents. The only diο¬erence is we need to follow our rules for fractions as well. It may be worth reviewing your notes on exponent properties to be sure your comfortable with using the properties. 312 8.6 Practice - Rational Exponents Write each expression in radical form. 3 5 1) m 3) (7x) 3 2 3 4 2) (10r)β 4) (6b) β 4 3 Write each expression in |
exponential form. 5) 7) 1 ( 6xβ )3 1 ( n4β )7 Evaluate. 2 3 9) 8 3 2 11) 4 6) vβ 8) 5aβ 1 4 10) 16 12) 100β 3 2 Simplify. Your answer should contain only positive exponents. 3 2 xy 1 2)β Β· 1 13) yx 1 3 1 2b 15) (a 17) a2b0 3a4 19) uv u Β· Β· 21) (x0y 1 3) 3 2)3 (v 3 2x0 23) a 7 4 b 3 4bβ1 Β· 3bβ1 25) β 5 4 3y β 1 3 yβ1 2y Β· 7 4 27) m 3 2nβ2 4 (mn 3)β1! 1 2)0 30) y0 3 4 yβ1) 1 3 (x 29) (m2n 3 4 n 31) (x y)ββ2 x 1 2 33) (uv2) β 1 4v2 v 32) 34) 313 1 vβ 2 3 14) 4v 16) (x 5 Β· 3 yβ 2)0 1 3 1 2 y 18) 2x 4 3 y 2x β 7 4 xy2)0 20) (x Β· 4v2 22) uβ 5 Β· 2xβ2y β 5 3 4 y 24) x β 5 1 3 26) ab 4a 2b Β· β 1 2b β 2 3 3 2 28) (y x β 1 2) 1 2 3 2 y xy Β· β 5 4 3 (u 2)β y0) β 4 3 (x β 2 3 xβ2y y4 Β· 1 3 y β2 y 5 β 3 3 y3) (x 2! 8.7 Radicals - Radicals of Mixed Index Objective: Reduce the index on a radical and multiply or divide radicals of diο¬erent index. Knowing that a radical has the same properties as exponents (written as a ratio) allows us to manipulate radicals in new ways. One thing we are allowed to do is reduce, not just the radicand, but the index as well. This is shown in the following example. Example 417. 8 x6y2 Rewrite as rational exponent (x6y2) p 1 5 Multiply exponents 6 2 x 8 y 8 Reduce each fraction |
3 1 4 All exponents have denominator of 4, this is our new index x 4 y x3y 4 Our Solution p What we have done is reduced our index by dividing the index and all the exponents by the same number (2 in the previous example). If we notice a common factor in the index and all the exponnets on every factor we can reduce by dividing by that common factor. This is shown in the next example Example 418. 24β a6b9c15 8β a2b3c5 Index and all exponents are divisible by 3 Our Solution We can use the same process when there are coeο¬cients in the problem. We will ο¬rst write the coeο¬cient as an exponential expression so we can divide the exponet by the common factor as well. Example 419. 9β 9β 8m6n3 23m6n3 3β 2m2n Write 8 as 23 Index and all exponents are divisible by 3 Our Solution We can use a very similar idea to also multiply radicals where the index does not match. First we will consider an example using rational exponents, then identify the pattern we can use. 314 Example 420. 3β ab2 4β a2b Rewrite as rational exponents (ab2) 1 3(a2b) 1 4 Multiply exponents To have one radical need a common denominator, 12 4 8 6 3 12 Write under a single radical with common index, 12 a 12 b 12β 12 b 12 a a4b8a6b3 12β a10b11 Add exponents Our Solution To combine the radicals we need a common index (just like the common denominator). We will get a common index by multiplying each index and exponent by an integer that will allow us to build up to that desired index. This process is shown in the next example. Example 421. 4β a2b3 6β a2b 12β a6b9a4b2 12β a10b11 Common index is 12. Multiply ο¬rst index and exponents by 3, second by 2 Add exponents Our Solution Often after combining radicals of mixed index we will need to simplify the resulting radical. Example 422. 5 x3y4 3 x2y p 15 p p x9y12x10y5 x19y17 15 x4 |
y2 p xy 15 Common index: 15. Multiply ο¬rst index and exponents by 3, second by 5 Add exponents Simplify by dividing exponents by index, remainder is left inside Our Solution p Just as with reducing the index, we will rewrite coeο¬cients as exponential expressions. This will also allow us to use exponent properties to simplify. Example 423. 3 4x2y 22x2y p 3 4 8xy3 23xy3 4 p p 12 p 12 p 28x8y429x3y9 217x11y13 12 25x11y 32x11y 2y p 2y 12 p Rewrite 4 as 22 and 8 as 23 Common index: 12. Multiply ο¬rst index and exponents by 4, second by 3 Add exponents (even on the 2) Simplify by dividing exponents by index, remainder is left inside Simplify 25 Our Solution p 315 If there is a binomial in the radical then we need to keep that binomial together through the entire problem. Example 424. 3 9x(y + z)2 3x(y + z) 3 32x(y + z)2 3x(y + z) p p 33x3(y + z)334x2(y + z)4 p 6 37x5(y + z)7 6 3x5(y + z) p 3(y + z) 6 p p Rewrite 9 as 32 Common index: 6. Multiply ο¬rst group by 3, second by 2 Add exponents, keep (y + z) as binomial Simplify, dividing exponent by index, remainder inside Our Solution p World View Note: Originally the radical was just a check mark with the rest of the radical expression in parenthesis. In 1637 Rene Descartes was the ο¬rst to put a line over the entire radical expression. The same process is used for dividing mixed index as with multiplying mixed index. The only diο¬erence is our ο¬nal answer cannot have a radical over the denominator. Example 425. 6 x4y3z2 8 p x7y2z p x16y12z8 x21y6z3 24 r Common index is 24. Multiply ο¬rst group by 4, second by 3 Subtract exponents 24 xβ 5y6z5 Negative exponent |
moves to denominator p y6z5 x5 24 r 24 y6z5 x5 r 24 x19 x19 r! Cannot have denominator in radical, need 12xβ²s, or 7 more Multiply numerator and denominator by 12β x7 x19y6z5 x 24 p Our Solution 316 8.7 Practice - Radicals of Mixed Index Reduce the following radicals. 8 1) 16x4y6 12 p3) 64x4y6z8 6 p5) q 12 7) 16x2 9y4 x6y9 8 p9) x6y4z2 9 p11) 8x3y6 p Combine the following radicals. 4 2) 9x2y6 4 p4) q 15 6) 25x3 16x5 x9y12z6 10 p8) 64x8y4 4 p10) 25y2 16 p12) 81x8y12 p 53β 6β xβ 7y3β 14) 16) 73β 54β 5β y3β 3z a2b3c 6 24) x2yz3 5 x2yz2 13) 15) 17) 3β xβ x 2 β xyβ 5 19) x2y 21) p 4 xy2 3 x2y 23) 25) 5β p 4β p a2bc2 4β aβ a3 27) 5β b3β b2 29) xy3 3 x2y 31) 4β p 9ab3 3a4bβ p 3 33) 3xy2z 4 9x3yz2 35) p 27a5(b + 1) p 3 81a(b + 1)4 37) 3β a2 p a4β p 39) 41) 43) 45) 4 x2y3 3β p xy ab3cβ a2b3cβ1 5β 4 (3x 5 p(3x β 1)3 1)3 p 3 β (2x + 1)2 5 p(2x + 1)2 p 18) 3x4β y + 4β 20) abβ 5β 2a2b2 22) 5β a2b3 4β a2b 26) 3β p x |
2 6β x5 p 28) 4β a3 3β a2 30) 5β a3b abβ 32) 2x3y3 3 4xy2 34) β p a4b3c4 3β p ab2c 8x (y + z)5 3 4x2(y + z)2 p 3β x2 p x5β 5β 3β a4b2 ab2 5 x3y4z9 p xyβ2z p 3 (2 + 5x)2 4 p (2 + 5x) 3x)3 3x)2 p 4 (5 3 p(5 p β β 36) 38) 40) 42) 44) 46) 317 8.8 Radicals - Complex Numbers Objective: Add, subtract, multiply, rationalize, and simplify expressions using complex numbers. World View Note: When mathematics was ο¬rst used, the primary purpose was for counting. Thus they did not originally use negatives, zero, fractions or irrational numbers. However, the ancient Egyptians quickly developed the need for βa partβ and so they made up a new type of number, the ratio or fraction. The Ancient Greeks did not believe in irrational numbers (people were killed for believing otherwise). The Mayans of Central America later made up the number zero when they found use for it as a placeholder. Ancient Chinese Mathematicians made up negative numbers when they found use for them. In mathematics, when the current number system does not provide the tools to solve the problems the culture is working with, we tend to make up new ways for dealing with the problem that can solve the problem. Throughout history this has been the case with the need for a number that is nothing (0), smaller than zero (negatives), between integers (fractions), and between fractions (irrational numbers). This is also the case for the square roots of negative numbers. To work with the square root of negative numbers mathematicians have deο¬ned what are called imaginary and complex numbers. Deο¬nition of Imaginary Numbers: i2 = 1 (thus i = 1β ) β β Examples of imaginary numbers include 3i, number is one that contains both a real and imaginary part, such as 2 + 5i. i and 3i 5β. A complex 6i, β 3 5 With this deο¬nition, the square root |
of a negative number is no longer undeο¬ned. We now are allowed to do basic operations with the square root of negatives. First we will consider exponents on imaginary numbers. We will do this by manipulating our deο¬nition of i2 = 1. If we multiply both sides of the deο¬nition β by i, the equation becomes i3 = i. Then if we multiply both sides of the equation again by i, the equation becomes i4 = 1) = 1, or simply i4 = 1. Multiplying again by i gives i5 = i. One more time gives i6 = i2 = 1. And if this pattern continues we see a cycle forming, the exponents on i change we cycle through simpliο¬ed answers of i, i, 1. As there are 4 diο¬erent possible β answers in this cycle, if we divide the exponent by 4 and consider the remainder, we can simplify any exponent on i by learning just the following four values: i2 = 1, β β β β β β ( Cyclic Property of Powers of i i0 = 1 i = i i2 = i3 = β β 1 i 318 Example 426. Example 427. i35 Divide exponent by 4 8R3 Use remainder as exponent on i i3 Simplify i Our Solution β i124 Divide exponent by 4 31R0 Use remainder as exponent on i i0 Simplify 1 Our Solution When performing operations (add, subtract, multilpy, divide) we can handle i just like we handle any other variable. This means when adding and subtracting complex numbers we simply add or combine like terms. Example 428. (2 + 5i) + (4 β 6 β 7i) Combine like terms 2 + 4 and 5i 2i Our Solution 7i β It is important to notice what operation we are doing. Students often see the parenthesis and think that means FOIL. We only use FOIL to multiply. This problem is an addition problem so we simply add the terms, or combine like terms. For subtraction problems the idea is the same, we need to remember to ο¬rst distribute the negative onto all the terms in the parentheses. Example 429. (4 8i) β 4 β 8i β (3 β 5i) Distribute the negative β 3 + 5i Combine like terms 4 1 3i Our Solution β 3 and β β 8i + 5i Addition and |
subtraction can be combined into one problem. Example 430. (5i) β (3 + 8i) + ( 8i 3 5i β β 4 + 7i) Distribute the negative 4 + 7i Combine like terms 5i 7 + 4i Our Solution β β β β 8i + 7i and 3 4 β β Multiplying with complex numbers is the same as multiplying with variables with one exception, we will want to simplify our ο¬nal answer so there are no exponents on i. 319 Example 431. Example 432. (3i)(7i) Multilpy coeο¬cients and iβ²s 21i2 Simplify i2 = 21( β β 1) Multiply 21 Our Solution 1 β 5i(3i 15i2 1) 15 Simplify i2 = 7) Distribute β 35i β 35i Multiply β 35i Our Solution β β β 1 β 15( Example 433. (2 6 + 10i β β 12i β 6 + 10i 6 + 10i 4i)(3 + 5i) 20i2 12i FOIL Simplify i2 = 1 β β 1) Multiply β 20( 12i + 20 Combine like terms 6 + 20 and 10i 26 2i Our Solution β β 12i β β Example 434. (3i)(6i)(2 18i2(2 36i2 3i) Multiply ο¬rst two monomials β 3i) Distribute β 54i3 β i) Multiply 54( 36 + 54i Our Solution Simplify i2 = 1 and i3 = β β β i 36( β 1) β β Remember when squaring a binomial we either have to FOIL or use our shortcut to square the ο¬rst, twice the product and square the last. The next example uses the shortcut Example 435. (4 5i)2 Use perfect square shortcut β 42 = 16 Square the ο¬rst 2(4)( β (5i)2 = 25i2 = 25( β 16 β 5i) = 1) = 40i 9 β β β β β 40i Twice the product 25 25 Combine like terms 40i Our Solution Square the last, simplify i2 = 1 β 320 β Dividing with complex numbers also has one thing we need to be careful of. If i is 1β, and it is in the denomin |
ator of a fraction, then we have a radical in the denominator! This means we will want to rationalize our denominator so there are no iβs. This is done the same way we rationalized denominators with square roots. Example 436. 7 + 3i 5i β 7 + 3i 5i β i i 7i + 3i2 5i2 β 7i + 3( 5( β 1) β 1) β Just a monomial in denominator, multiply by i Distribute i in numerator Simplify i2 = 1 β Multiply 7i 3 β 5 Our Solution The solution for these problems can be written several diο¬erent ways, for example i, The author has elected to use the solution as written, but it is β important to express your answer in the form your instructor prefers. 5 + 7 3 + 7i 5 or β 3 5 Example 437. 6i 2 β 4 + 8i Binomial in denominator, multiply by conjugate, 4 8i β 6i 2 β 4 + 8i 8i 8i 4 4 β β FOIL in numerator, denominator is a diο¬erence of squares 8 β 16i 16 24i + 48i2 64i2 β β Simplify i2 = 1 β 8 β 16i 16 24i + 48( 1) 64( β β β 1) β Multiply 8 β 16i 24i β 16 + 64 β 48 Combine like terms 8 48 and 16i β β β 24i and 16 + 64 40i β 40 β 80 β i 1 β 2 Reduce, divide each term by 40 Our Solution 321 Using i we can simplify radicals with negatives under the root. We will use the product rule and simplify the negative as a factor of negative one. This is shown in the following examples. Example 438. β Example 439. 16β β 16 1 Β· 4i Our Solution β Consider the negative as a factor of Take each root, square root of β 1 is i 1 β 24β β 6 4 1 Β· Β· 2i 6β β β Find perfect square factors, including Square root of Our Solution 1 1 is i, square root of 4 is 2 β β When simplifying complex radicals it is important that we take the radical (as an i) before we combine radicals. β 1 out of the Example 440. Simplify the negatives, bringing i out of radicals 1, also multiply radicals 3οΏ½ |
οΏ½ β 6β β (i 6β )(i 3β ) Multiply i by i is i2 = Simplify the radical Take square root of 9 Our Solution 18β β 2β 9 Β· 3 2β β β β If there are fractions, we need to make sure to reduce each term by the same number. This is shown in the following example. Example 441. 15 β 200 Simplify the radical ο¬rst β β β 20 β β 1 β Β· 15 β β β 20 3 200 β 100 2 Β· 10i 2β 10i 2β 2i 2β β 4 Find perfect square factors, including Take square root of Put this back into the expression 1 and 100 β 1 β All the factors are divisible by 5 Our Solution By using i = we will be able to simplify and solve problems that we could not simplify and solve before. This will be explored in more detail in a later section. 1β β 322 8.8 Practice - Complex Numbers Simplify. 8 + 4i) ( β 2) (3i) (7i) β 1) 3 β 3) (7i) 5) ( β 7) (3 β 9) (i) β 11) (6i)( 13) ( β (3 2i) β (3 + 7i) β 6i) β 3i) + ( β (2 + 3i) 7 8i) β 6 β 8i) β 5i)(8i) 4) 5 + ( 6 6i) β 8i) β (7i) (5 3i) β β i) + (1 β 5i) β 4i) + (8 β β 4i) 6) ( β 8) ( β 10) (5 4 β 12) (3i)( 14) (8i)( 8i) 4i) β β 7i)2 15) ( β 17) (6 + 5i)2 19) ( 21) ( β β 7 4i)( 8 + 6i) β β 4 + 5i)(2 7i) β 4 + 2i) 23) ( 8 6i)( β β β 25) (1 + 5i)(2 + i) 27) β 9 + 5i i 29) β 9i 10 6i β 31) β 6i 3 β 4i i 33) 10 β β i 35) 37) 39) 41 |
) 4i 10 + i β 8 6i 7 β 7 β 7i 10 5i 6 β i β 16) ( β i)(7i)(4 3i) β 18) (8i)( β 20) (3i)( 2i)( β 3i)(4 2 8i) β 4i) β β 8(4 22) β 24) ( β 26) ( β 6i)(3 β 2 + i)(3 8i) 2( 2 6i) β 2i) β β (7i)(4i) β 5i) β β 3 + 2i 3i β 28) β 30) β 32) β 4 + 2i 3i 5 + 9i 9i 34) 10 5i 36) 38) 40) 42) 9i 5i 1 β 4 4 + 6i 9 8 β β 6i 8i 7i 6 β 323 43) 81β β 45) 10β β 2β β 47 3 + 27β β 6 49) 8 16β β β 4 51) i73 53) i48 55) i62 57) i154 44) 45β β 46) 12β β 2β β 48) β 4 8β β β 4 β 50) 6 + 32β β 4 52) i251 54) i68 56) i181 58) i51 324 Chapter 9 : Quadratics 9.1 Solving with Radicals.............................................................................326 9.2 Solving with Exponents..........................................................................332 9.3 Complete the Square..............................................................................337 9.4 Quadratic Formula.................................................................................343 9.5 Build Quadratics From Roots................................................................348 9.6 Quadratic in Form.................................................................................352 9.7 Application: Rectangles.........................................................................357 9.8 Application: Teamwork..........................................................................364 9.9 Simultaneous Products...........................................................................370 9.10 Application: Revenue and Distance......................................................373 9.11 Graphs of Quadratics............................................................................380 325 9.1 Quadratics - Solving with Radicals Objective: Solve equations with radicals and check for extraneous solutions. Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root we can rise both sides to the second power. To clear a cubed root we can raise both |
sides to a third power. There is one catch to solving a problem with roots in it, sometimes we end up with solutions that do not actually work in the equation. This will only happen if the index on the root is even, and it will not happen all the time. So for these problems it will be required that we check our answer in the original problem. If a value does not work it is called an extraneous solution and not included in the ο¬nal solution. When solving a radical problem with an even index: check answers! Example 442. 7x + 2 β β ( 7x + 2 )2 = 42 7x + 2 = 16 2 2 = 4 Even index! We will have to check answers Square both sides, simplify exponents Solve Subtract 2 from both sides β β 7x = 14 Divide both sides by 7 7 7 x = 2 Need to check answer in original problem 7(2) + 2 β 14 + 2 = 4 Multiply = 4 Add p 16β = 4 Square root 4 = 4 True! It works! x = 2 Our Solution Example 443. 3β 3β ( x β = x 1 β )3 = ( 1 = 1 x 4 Odd index, we donβ²t need to check answer β 4)3 Cube both sides, simplify exponents 64 Solve β β β 326 + 1 + 1 Add 1 to both sides 63 Our Solution x = β Example 444. 4β 4β ( 3x + 6 = 3x + 6 ) = ( 3x + 6 = 81 6 6 β 3 Even index! We will have to check answers β 3)4 Rise both sides to fourth power Solve Subtract 6 from both sides 3 β β 3x = 75 Divide both sides by 3 3 x = 25 Need to check answer in original problem = = 814β = 3 = 3 Multiply 3 Add 3 Take root 3 False, extraneous solution β β β β No Solution Our Solution 4 p 3(25) + 6 4β 75 + 6 If the radical is not alone on one side of the equation we will have to solve for the radical before we raise it to an exponent Example 445. β x + 4x + 1 x β 4x + 1 = 5 β = 5 Even index! We will have to check solutions x x Isolate radical by subtracting x from both sides Square both sides )2 = (5 β β β |
x)2 Evaluate exponents, recal (a order terms β β β β β β 12)(x 10x + x2 Re 10x + 25 Make equation equal zero 4x 1 14x + 24 2 Subtract 4x and 1 from both sides Factor Set each factor equal to zero Solve each equation β ( 4x + 1 4x + 1 = 25 4x + 1 = x2 4x 1 0 = x2 β β 0 = (x 12 = 0 or x β x β + 12 + 12 b)2 = a2 β β 2ab + b2 x = 12 or x = 2 Need to check answers in original problem (12) + 4(12) + 1 = 5 Check x = 5 ο¬rst p 327 = 5 Add β 12 + 48 + 1 12 + 49β = 5 Take root 12 + 7 = 5 Add 19 = 5 False, extraneous root (2) + 4(2) + 1 2 + 8 + 1β p = 5 Check x = 2 = 5 Add 2 + 9β = 5 Take root 2 + 3 = 5 Add 5 = 5 True! It works x = 2 Our Solution The above example illustrates that as we solve we could end up with an x2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index in the problem was even. Sometimes both values work, sometimes only one, and sometimes neither works. World View Note: The babylonians were the ο¬rst known culture to solve quadratics in radicals - as early as 2000 BC! If there is more than one square root in a problem we will clear the roots one at a time. This means we must ο¬rst isolate one of them before we square both sides. Example 446. β 3x xβ = 0 Even index! We will have to check answers Isolate ο¬rst root by adding xβ to both sides Square both sides β β 8 + xβ + xβ = xβ 3x 8 8 β β β ( 3x β )2 = ( xβ )2 Evaluate exponents 3x Solve Subtract 3x from both sides 3x 8 = x 3x 2x Divide both sides by β Need to check answer in original 3(4) β 12 p 8 |
β 8 β 4β β β β 4β = 0 Multiply 4β = 0 Subtract 4β = 0 Take roots 328 2 β Subtract 2 = 0 0 = 0 True! It works x = 4 Our Solution When there is more than one square root in the problem, after isolating one root and squaring both sides we may still have a root remaining in the problem. In this case we will again isolate the term with the second root and square both sides. When isolating, we will isolate the term with the square root. This means the square root can be multiplied by a number after isolating. Example 447. β 2x + 1 xβ = 1 Even index! We will have to check answers β + xβ + xβ = xβ + 1 β 2x + 1 Isolate ο¬rst root by adding xβ to both sides Square both sides β ( 2x + 1 )2 = ( xβ + 1)2 Evaluate exponents, recall (a + b)2 = a2 + 2ab + b2 2x + 1 = x + 2 xβ + 1 x 1 x = 2 xβ β β β β x 1 Isolate the term with the root Subtract x and 1 from both sides Square both sides (x)2 = (2 xβ )2 Evaluate exponents β x2 x(x x = 0 or x x2 = 4x Make equation equal zero Subtract x from both sides 4x Factor Set each factor equal to zero Solve β β β + 4 + 4 Add 4 to both sides of second equation 4x β 4x = 0 4) = 0 4 = 0 x = 0 or x = 4 Need to check answers in original 2(0) + 1 β 1β p 2(4) + 1 β 8 + 1β 9β p = 1 Check x = 0 ο¬rst (0) 0β = 1 Take roots β Subtract 0 = 1 1 = 1 True! It works Check x = 4 (4) 4β = 1 Add 4β = 1 Take roots Subtract 2 = 1 1 = 1 True! It works 329 x = 0 or 4 Our Solution Example 448. β 3x + 9 x + 4β β + x + 4β 3x + 9 β β = + |
x + 4β 1 1 Even index! We will have to check answers Isolate the ο¬rst root by adding x + 4β Square both sides = x + 4β β ( 3x + 9 3x + 9 = x + 4 )2 = ( x + 4β β 2 x + 4β β 3x + 9 = x + 5 x x 5 5 β 2x + 4 = β β β 2 x + 4β β 2 x + 4β β 1)2 Evaluate exponents + 1 Combine like terms Isolate the term with radical Subtract x and 5 from both sides Square both sides β (2x + 4)2 = ( 2 x + 4β 4x2 + 16x + 16 = 4(x + 4) Distribute 4x2 + 16x + 16 = 4x + 16 Make equation equal zero )2 Evaluate exponents β β β β β 4x 4x 16 16 4x2 + 12x = 0 4x(x + 3) = 0 4x = 0 or x + 3 = 0 4 3 3 Check solutions in original Subtract 4x and 16 from both sides Factor Set each factor equal to zero Solve 3 x = 0 or x = β 4 β β 3(0) + 9 β p 9β p β 3 (0) + 4 = 4β = 2 = 1 = β 3( 3β β β p p 3) + 4 = 3) + 4 = 1 Check x = 0 ο¬rst 1 Take roots 1 1 Subtract False, extraneous solution 3 β 1 Check x = 1 Add 1 Take roots Subtract 1 1 True! It works 3 Our Solution 330 9.1 Practice - Solving with Radicals Solve. 1) β 2x + 3 3) β 6x ) 3 + x = 6x + 13 β 7) β 3 3x β 9) β 4x + 5 β β 1 = 2x x + 4β = 2 11) β 2x + 4 13) β 2x + 6 15) β 6 2x β β β β x + 3β = 1 x + 4β = 1 β 2x + 3 = 3 2) β 5x + 1 4 = 0 β 4) x + 2β xβ = 2 β 6) x 1 = 7 |
xβ β β 8) β 2x + 2 = 3 + 2x β 1 β 10) β 3x + 4 12) β 7x + 2 14) β 4x 3 β 16) β 2 3x β β β β β x + 2β = 2 β 3x + 6 = 6 β 3x + 1 = 1 β 3x + 7 = 3 331 9.2 Quadratics - Solving with Exponents Objective: Solve equations with exponents using the odd root property and the even root property. Another type of equation we can solve is one with exponents. As you might expect we can clear exponents by using roots. This is done with very few unexpected results when the exponent is odd. We solve these problems very straight forward using the odd root property Odd Root Property: if an = b, then a = bnβ when n is odd Example 449. 5β = 325β x5 x5 = 32 Use odd root property Simplify roots x = 2 Our Solution However, when the exponent is even we will have two results from taking an even root of both sides. One will be positive and one will be negative. This is because both 32 = 9 and ( 3)2 = 9. so when solving x2 = 9 we will have two solutions, one β positive and one negative: x = 3 and 3 β Even Root Property: if an = b, then a = bnβ when n is even Β± Example 450. x4 = 16 Use even root property ( ) Β± 332 4β = x4 164β Β± x = Β± Simplify roots 2 Our Solution World View Note: In 1545, French Mathematicain Gerolamo Cardano published his book The Great Art, or the Rules of Algebra which included the solution of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions! Example 451. (2x + 4)2 = 36 Use even root property ( ) Β± (2x + 4)2 = Β± 2x + 4 = 2x + 4 = 6 or 2x + 4 = p 4 β 4 4 β 2x = 2 or 2x = 2 β 2 2 x = 1 or x = β 36β Simplify roots Subtract 4 from both sides 6 To avoid sign errors we need two equations Β± 6 One equation for + |
, one equation for β 4 β 10 Divide both sides by 2 β 2 5 Our Solutions β In the previous example we needed two equations to simplify because when we 6. If the roots did took the root, our solutions were two rational numbers, 6 and not simplify to rational numbers we can keep the in the equation. β Β± Example 452. (6x β 6x p ) Β± 45β 3 5β 9)2 = 45 Use even root property ( (6x β 9)2 Simplify roots = Β± Use one equation because root did not simplify to rational 9 = β Β± + 9 + 9 Add 9 to both sides 6x = 9 Divide both sides by 6 6 Simplify, divide each term by 3 3 5β Β± 6 3 5β Β± 2 Our Solution 333 When solving with exponents, it is important to ο¬rst isolate the part with the exponent before taking any roots. Example 453. (x + 4)3 6 = 119 β + 6 + 6 Isolate part with exponent (x + 4)3 = 125 Use odd root property (x + 4)3 = 125 Our Solution Simplify roots Solve Subtract 4 from both sides β β 3 p Example 454. (6x + 1)2 + 6 = 10 6 6 β β Isolate part with exponent Subtract 6 from both sides (6x + 1)2 = 4 Use even root property ( ) Β± 4β Simplify roots (6x + 1)2 = Β± 6x + 1 = 6x + 1 = 2 or 6x + 1 = p 1 β 1 β 1 β 6x = 1 or 6x = 6 6 x = or x = 6 1 6 Solve each equation Subtract 1 from both sides 2 To avoid sign errors, we need two equations Β± 2 β 1 β 3 Divide both sides by 6 β 6 1 2 Our Solution β When our exponents are a fraction we will need to ο¬rst convert the fractional exponent into a radical expression to solve. Recall that a. Once we have done this we can clear the exponent using either the even ( ) or odd root property. Then we can clear the radical by raising both sides to an exponent (remember to check answers if the index is even). n = ( anβ ) Β± m m Example 455. 2 (4x + 1) 5β ( 4x + 1 5 = 9 Rewrite as a radical expression )2 = |
9 Clear exponent ο¬rst with even root property ( 5β ( 4x + 1 )2 = 9β Β± q Simplify roots 334 ) Β± = 3 Clear radical by raising both sides to 5th power 5β 5β ( 4x + 1 4x + 1 )5 = ( 4x + 1 = 4x + 1 = 243 or 4x + 1 = 1 1 1 Β± 3)5 243 243 1 Β± Β± β β β β 4x = 242 or 4x = 4 4 β 4 121 2 x = Simplify exponents Solve, need 2 equations! Subtract 1 from both sides β 244 Divide both sides by 4 4 61 Our Solution, β Example 456. 3 β β )3 2 3 q β 4β 4β ( 3x 3x (3x 4β ( 3x 2) 2 4 = 64 Rewrite as radical expression )3 = 64 Clear exponent ο¬rst with odd root property 4β ( 3x 3β = 64 Simplify roots 3x = 4 Even Index! Check answers. 2 β )4 = 44 Raise both sides to 4th power 2 2 = 256 β Solve β + 2 + 2 Add 2 to both sides 3x = 258 Divide both sides by 3 3 3 x = 86 Need to check answer in radical form of problem )3 = 64 Multiply )3 = 64 Subtract 4 ( 3(86) 4β ( 258 2 β 2 β 4β )3 = 64 Evaluate root ( 256 p 43 = 64 Evaluate exponent 64 = 64 True! It works x = 86 Our Solution With rational exponents it is very helpful to convert to radical form to be able to see if we need a because we used the even root property, or to see if we need to check our answer because there was an even root in the problem. When checking we will usually want to check in the radical form as it will be easier to evaluate. Β± 335 9.2 Practice - Solving with Exponents Solve. 1) x2 = 75 3) x2 + 5 = 13 5) 3x2 + 1 = 73 7) (x + 2)5 = 243 β 9) (2x + 5)3 6 = 21 β 11) (x 13) (2 2 1) 3 = 16 3 x) 2 = 27 β β 15) (2x β 2 3) 3 = 4 17 |
) (x + 1 2 )β 2 3 = 4 19) (x β 5 1)β 2 = 32 21) (3x β 4 2) 5 = 16 23) (4x + 2) 5 = 3 8 β 2) x3 = 8 β 4) 4x3 6) (x β 2 = 106 β 4)2 = 49 8) (5x + 1)4 = 16 10) (2x + 1)2 + 3 = 21 12) (x β 3 1) 2 = 8 14) (2x + 3) 3 = 16 4 16) (x + 3)β 3 = 4 1 18) (x β 5 1)β 3 = 32 3 20) (x + 3) 2 = 8 β 22) (2x + 3) 2 = 27 3 24) (3 β 4 2x) 3 = 81 β 336 9.3 Quadratics - Complete the Square Objective: Solve quadratic equations by completing the square. When solving quadratic equations in the past we have used factoring to solve for our variable. This is exactly what is done in the next example. Example 457. x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x + 3 = 0 or x + 2 = 0 2 2 Our Solutions Factor Set each factor equal to zero Solve each equation 2 β x = β x = β 3 or 3 3 β β β 2x However, the problem with factoring is all equations cannot be factored. Consider the following equation: x2 7 = 0. The equation cannot be factored, however β there are two solutions to this equation, 1 + 2 2β and 1 2 2β. To ο¬nd these two solutions we will use a method known as completing the square. When completing the square we will change the quadratic into a perfect square which can easily be solved with the square root property. The next example reviews the square root property. β β Example 458. p (x + 5)2 (x + 5)2 = 18 18β 3 2 β Β± Square root of both sides Simplify each radical Subtract 5 from both sides 3 2β Our Solution 337 To complete the square, or make our problem into the form of the previous example, we will be searching for the third term in a trinomial. If a quadratic is of the form x2 + b x + c |
, and a perfect square, the third term, c, can be easily found by the formula 2. This is shown in the following examples, where we b ο¬nd the number that completes the square and then factor the perfect square. 1 2 Β· Example 459. x2 + 8x + c c = 2 1 2 Β· b 2 and our b = 8 8 = 42 = 16 The third term to complete the square is 16 x2 + 8x + 16 Our equation as a perfect square, factor (x + 4)2 Our Solution 1 2 Β· Example 460. x2 7x + c c = 1 2 Β· b 2 and our = 49 4 49 4 2 x2 β 11x + 7 2 x β The third term to complete the square is Our equation as a perfect square, factor Our Solution Example 461. x2 + and our b = 8 = 25 36 The third term to complete the square is 338 49 4 25 36 x2 + 5 3 x + x + 5 6 25 36 2 Our equation as a perfect square, factor Our Solution The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following ο¬ve steps describe the process used to complete the square, along with an example to demonstrate each step. Problem 3x2 + 18x 1. Separate constant term from variables 2. Divide each term by a 3. Find value to complete the square: b 1 2 Β· 2 4. Add to both sides of equation 5. Factor Solve by even root property = 32 = 9 3x2 + 18x x2 + 18 3 3 3 x2 + 6x 2 x 6 1 2 Β· x2 + 6x = 2 + 9 + 9 x2 + 6x + 9 = 11 (x + 3)2 = 11 (x + 3)2 = Β± 11β Β± 3 β 11 = β 11β World View Note: The Chinese in 200 BC were the ο¬rst known culture group to use a method similar to completing the square, but their method was only used to calculate positive roots. The advantage of this method is it can be used to solve any quadratic equation. The following examples show how completing the square can give us rational solutions, irrational solutions, and even complex solutions. Example 462. 2x2 + 20x + 48 = 0 Separate constant term from varaibles 339 β 2x2 + 20x |
2 2 48 = β β 48 Subtract 24 48 Divide by a or 2 2 x2 + 10x 24 Find number to complete the square: = β 2 1 2 Β· b 2 10 = 52 = 25 Add 25 to both sides of the equation 1 2 Β· x2 + 10x β = 24 + 25 + 25 x2 + 10x + 25 = 1 (x + 5)2 = 1 1x + 5)2 Β± 5 p Factor Solve with even root property Simplify roots Subtract 5 from both sides β x = β 5 β 4 or 1 Evaluate 6 Our Solution Β± β x = β Example 463. x2 3x β 2 = 0 Separate constant from variables β + 2 + 2 Add 2 to both sides 3x β = 2 No a, ο¬nd number to complete the square 2 = 9 4 Add 9 4 to both sides, 1 2 Β· b 2 x2 x2 β 3x + = = = 17 4 17 4 17 4 Need common denominator (4) on right Factor Solve using the even root property 3 2 x β s 2 = Β± = Β± 17 4 r 17β 2 3 2 β x Simplify roots Add 3 2 to both sides, 340 + 3 2 + 3 2 we already have a common denominator x = 3 Β± 17β 2 Our Solution Example 464. x2 1 2 Β· 2 3 3x2 = 2x 2x 2x β 2x = 3 β β 3x2 β Separate the constant from the variables Subtract 2x from both sides 7 Divide each term by a or 3 β 3 7 3 β = 1 9 Add to both sides, Find the number to complete the square 21 = β 3 + 1 9 20 = β 9 get common denominator on right x2 β 2 3 x + = β 1 3 2 20 9 20 9 Factor Solve using the even root property 20 β 9 Β± r Simplify roots + 2i 5β 3 Add 1 3 to both sides, Already have common denominator x = 1 Β± 2i 5β 3 Our Solution As several of the examples have shown, when solving by completing the square we will often need to use fractions and be comfortable ο¬nding common denominators and adding fractions together. Once we get comfortable solving by completing the square and using the ο¬ve steps, any quadratic equation can be easily solved. 341 9.3 Practice - Complete the Square Find the value that |
completes the square and then rewrite as a perfect square. 1) x2 3) m2 5) x2 7) y2 β β β β 30x + __ 36m + __ 15x + __ y + __ 2) a2 4) x2 6) r2 8) p2 β β β β 24a + __ 34x + __ 1 9 17p + __ r + __ β 14 = 4 6 β β β β β β β β β β β β β β β 29 16p 12n 24 = 3 = 6 1 = 0 52 = 0 57 = 4 β 10x 6x + 47 = 0 8v + 45 = 0 β 37 = 5 10k + 48 = 0 16x + 55 = 0 16n + 67 = 4 β 21 + 10n 10) n2 8n 12 = 0 12) b2 + 2b + 43 = 0 14) 3x2 16) 8a2 + 16a 18) p2 20) m2 8m 22) 6r2 + 12r 24) 6n2 β β 26) v2 = 14v + 36 28) a2 β 30) 5x2 = 32) 5n2 = β 34) x2 + 8x + 15 = 8 36) n2 + 4n = 12 Solve each equation by completing the square. 9) x2 β 11) v2 13) 6x2 + 12x + 63 = 0 15) 5k2 17) x2 + 10x 19) n2 21) 2x2 + 4x + 38 = 23) 8b2 + 16b 25) x2 = 27) n2 = 29) 3k2 + 9 = 6k 31) 2x2 + 63 = 8x 33) p2 β 35) 7n2 37) 13b2 + 15b + 44 = 39) 5x2 + 5x = 31 41) v2 + 5v + 28 = 0 43) 7x2 45) k2 β 47) 5x2 + 8x 49) m2 = 51) 8r2 + 10r = 53) 5n2 38) 40) 8n2 + 16n = 64 42) b2 + 7b 44) 4x2 + 4x + 25 = 0 46) a2 β 48) 2p2 50) n2 β 52) 3x2 54) 4b2 β 56) 10v2 8p = 55 n + |
7 = 7n + 6n2 β 7k + 50 = 3 β 2x2 + 3x β 15 + 9m 3n + 6 + 4n2 5 + 7b2 + 3b 6x + 40 = 0 5a + 25 = 3 β 11x = 8n + 60 = β n = 10n + 15 26 + 10x p + 56 = β 5 = β 5x 33 = 0 40 = 8 56 = 10a 4x2 41 55 18 β β β β β β β β β β β β 55) β β β 8 3r2 + 12r + 49 = 6 6r2 β β 15b + 56 = 3b2 15v = 27 + 4v2 6v β β 342 9.4 Quadratics - Quadratic Formula Objective: Solve quadratic equations by using the quadratic formula. The general from of a quadratic is ax2 + bx + c = 0. We will now solve this formula for x by completing the square Example 465. 1 2 Β· b a c Separate constant from variables Subtract c from both sides ax2 + bc + c = 0 c β c Divide each term by b2 4a2 Add to both sides, b 2a = ax2 + bx a a b a x2 + 2 = Find the number that completes the square b2 4a2 β c a 4a 4a = b2 4a2 β 4ac 4a2 = b2 4ac β 4a2 Get common denominator on right x2 + b a x + b2 4a2 = b2 4a2 β 4ac 4a2 = b2 4ac β 4a2 Factor x + b 2a 2 = b2 4ac β 4a2 Solve using the even root property x + b 2a s x + b 2a 2 = b2 4ac β 4a2 Β± r Simplify roots = Β± 4ac β b2 β 2a Subtract b 2a from both sides x = β b Β± β b2 2a β 4ac Our Solution This solution is a very important one to us. As we solved a general equation by completing the square, we can use this formula to solve any quadratic equation. Once we identify what a, b, and c are in the quadratic, we can substitute those 343 values into x = β known as the quadratic fromula β |
Β± b b2 p2a 4ac and we will get our two solutions. This formula is Quadratic Formula: if ax2 + b x + c = 0 then x = β b Β± b2 p2a 4ac β World View Note: Indian mathematician Brahmagupta gave the ο¬rst explicit formula for solving quadratics in 628. However, at that time mathematics was not done with variables and symbols, so the formula he gave was, βTo the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divided by twice the to square 4ac + b2 2a as the solution to the equation ax2 + bx = c. translate value.β would This the is β b p We can use the quadratic formula to solve any quadratic, this is shown in the following examples. Example 466. x = β 3 Β± x2 + 3x + 2 = 0 4(1)(2) 32 β p2(1β 9 β 2 1β 3 Β± 2 3 Β± 2 4 or β or β β a = 1, b = 3, c = 2, use quadratic formula Evaluate exponent and multiplication Evaluate subtraction under root Evaluate root Evaluate Β± to get two answers Simplify fractions 2 Our Solution As we are solving using the quadratic formula, it is important to remember the equation must ο¬st be equal to zero. Example 467. 25x2 = 30x + 11 30x 11 11 β 11 = 0 11) β 30x β 4(25)( β 30x β 25x2 30)2 β β 2(25) β 30 Β± x = β ( p First set equal to zero Subtract 30x and 11 from both sides a = 25, b = 30, c = β β 11, use quadratic formula Evaluate exponent and multiplication 344 x = β 30 Β± 900 + 1100 50 x = 30 2000β 50 Β± x = 30 20 5β 50 Β± x = 3 Β± 2 5β 5 Evaluate addition inside root Simplify root Reduce fraction by dividing each term by 10 Our Solution Example 468. 3x2 + 4x + 8 = 2x2 + 6x 2x2 x2 2x2 β β 5 β 6x + 5 β β β 2 6x + 5 2x + 13 |
= 0 2)2 β β p 2(1(1)(13) β 52 β 4 2 48β Β± β 2 4i 3β 2 2i 3 First set equation equal to zero Subtract 2x2 and 6x and add 5 a = 1, b = 2, c = 13, use quadratic formula β Evaluate exponent and multiplication Evaluate subtraction inside root Simplify root Reduce fraction by dividing each term by 2 Our Solution When we use the quadratic formula we donβt necessarily get two unique answers. We can end up with only one solution if the square root simpliο¬es to zero. Example 469. 12 x = Β± 4x2 ( β 12)2 β p 2(4) β 12 x = Β± 12x + 9 = 0 4(4)(9) a = 4, b = β 12, c = 9, use quadratic formula Evaluate exponents and multiplication 144 β Evaluate subtraction inside root 0β Evaluate root β 144 8 12 x = x = Β± 8 12 0 Β± 8 12 x = 8 3 2 x = Evaluate Β± Reduce fraction Our Solution 345 If a term is missing from the quadratic, we can still solve with the quadratic formula, we simply use zero for that term. The order is important, so if the term with x is missing, we have b = 0, if the constant term is missing, we have c = 0. Example 470. x = β 0 Β± 3x2 + 7 = 0 02 4(3)(7) β p2(3) 84β x = Β± β 6 2i 21β 6 i 21, b = 0(missing term), c = 7 Evaluate exponnets and multiplication, zeros not needed Simplify root Reduce, dividing by 2 Our Solution We have covered three diο¬erent methods to use to solve a quadratic: factoring, complete the square, and the quadratic formula. It is important to be familiar with all three as each has its advantage to solving quadratics. The following table walks through a suggested process to decide which method would be best to use for solving a problem. 1. If it can easily factor, solve by factoring 2. If a = 1 and b is even, complete the square 3. Otherwise, solve by the quadratic formula β β x2 5x + 6 = 0 (x 2 |
)(x x = 2 or x = 3 x2 + 2x = 4 β 3) = 0 2 2 = 12 = 1 1 2 Β· 5β 5β x2 + 2x + 1 = 5 (x + 1) x2 β x = 3 x = 3 Β± β Β± 3x + 4 = 0 ( 3)2 β p 2(1) i 7β 2 Β± Β± β 4(1)(4) The above table is mearly a suggestion for deciding how to solve a quadtratic. Remember completing the square and quadratic formula will always work to solve any quadratic. Factoring only woks if the equation can be factored. 346 9.4 Practice - Quadratic Formula Solve each equation with the quadratic formula. 1) 4a2 + 6 = 0 3) 2x2 8x β β 2 = 0 5) 2m2 3 = 0 β 7) 3r2 9) 4n2 11) v2 β β β 2r β 1 = 0 36 = 0 4v 5 = 8 β β 13) 2a2 + 3a + 14 = 6 2) 3k2 + 2 = 0 4) 6n2 1 = 0 β 6) 5p2 + 2p + 6 = 0 8) 2x2 2x 15 = 0 β β 10) 3b2 + 6 = 0 12) 2x2 + 4x + 12 = 8 14) 6n2 3n + 3 = 4 β β 16) 4x2 14 = 2 β β 18) 4n2 + 5n = 7 20) m2 + 4m 48 = 3 β β 3 = 8b 22) 3b2 β 24) 3r2 + 4 = 6r β 26) 6a2 = 5a + 13 β 28) 6v2 = 4 + 6v 30) x2 = 8 32) 2k2 + 6k 16 = 2k β 34) 12x2 + x + 7 = 5x2 + 5x 4 = 7 16 = 2 β 16 = 4 15) 3k2 + 3k 17) 7x2 + 3x β β 19) 2p2 + 6p β 21) 3n2 + 3n = 3 β 7x + 49 23) 2x2 = β 25) 5x2 = 7x + 7 27) 8n2 = 3n β 29) 2x2 + 5x = 8 |
3 β β 31) 4a2 64 = 0 β 33) 4p2 + 5p 36 = 3p2 β 3n 35) β 37) 7r2 39) 2n2 5n2 β 52 = 2 7n2 β β 36) 7m2 6m + 6 = m β β 12 = 3r β 9 = 4 β β 38) 3x2 3 = x2 β 40) 6b2 = b2 + 7 b β 347 9.5 Quadratics - Build Quadratics From Roots Objective: Find a quadratic equation that has given roots using reverse factoring and reverse completing the square. Up to this point we have found the solutions to quadratics by a method such as factoring or completing the square. Here we will take our solutions and work backwards to ο¬nd what quadratic goes with the solutions. We will start with rational solutions. If we have rational solutions we can use factoring in reverse, we will set each solution equal to x and then make the equation equal to zero by adding or subtracting. Once we have done this our expressions will become the factors of the quadratic. Example 471. The solutions are 4 and x = 4 or x = 4 4 + 2 + 2 β β Set each solution equal to x 2 2 Make each equation equal zero Subtract 4 from ο¬rst, add 2 to second x β β 4 = 0 or x + 2 = 0 These expressions are the factors (x 4)(x + 2) = 0 FOIL β β x2 + 2x x2 β 2x β β β 4x 8 Combine like terms 8 = 0 Our Solution If one or both of the solutions are fractions we will clear the fractions by multiplying by the denominators. Example 472. The solution are and 2 3 or Set each solution equal to x Clear fractions by multiplying by denominators 3x 2 β 3x = 2 or 4x = 3 Make each equation equal zero 2 3 3 = 0 These expressions are the factors 3) = 0 FOIL 8x + 6 = 0 Combine like terms β 2 = 0 or 4x 2)(4x (3x β 12x2 9x β β β β β 3 β β Subtract 2 from the ο¬rst, subtract 3 from the second 348 12x2 β 17x + 6 = 0 Our Solution If the solutions have radicals (or complex numbers) then we cannot use reverse factoring. |
In these cases we will use reverse completing the square. When there are radicals the solutions will always come in pairs, one with a plus, one with a. We will then set this minus, that can be combined into βoneβ solution using solution equal to x and square both sides. This will clear the radical from our problem. Β± Example 473. The solutions are 3β and x = 3β Write as β²β²oneβ²β² expression equal to x 3β β Square both sides Β± x2 = 3 Make equal to zero 3 3 = 0 Our Solution β 3 Subtract 3 from both sides x2 β β We may have to isolate the term with the square root (with plus or minus) by adding or subtracting. With these problems, remember to square a binomial we use the formula (a + b)2 = a2 + 2ab + b2 Example 474. The solutions are 2 β 5 2β and 2 + 5 2β Write as β²β²oneβ²β² expression equal to x Isolate the square root term Subtract 2 from both sides Square both sides Β± x = 2 2 2 β β x 2 = β 4x + 4 = 25 Β± 5 2β 5 2β 2 Β· x2 β x2 β 4x + 4 = 50 Make equal to zero 50 50 Subtract 50 46 = 0 Our Solution x2 β β 4x β β World View Note: Before the quadratic formula, before completing the square, before factoring, quadratics were solved geometrically by the Greeks as early as 300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations geometrically! If the solution is a fraction we will clear it just as before by multiplying by the denominator. 349 Example 475. The solutions are 2 + 3β 4 2 2 and x = 3β β 4 3β Β± 4 3β Write as β²β²oneβ²β² expresion equal to x Clear fraction by multiplying by 4 Β± 4x = 2 2 2 β 2 = Isolate the square root term Subtract 2 from both sides Square both sides β 16x + 4 = 3 Make equal to zero 3β Β± β 4x 3 3 Subtract 3 β β 16x + 1 = 0 Our Solution 16x2 16x2 β β The process used for complex solutions is identical to the process used for radicals. Example 4 |
76. The solutions are 4 5i and 4 + 5i Write as β²β²oneβ²β² expression equal to x β Β± 5i β x x = 4 4 4 β 5i 4 = 8x + 16 = 25i2 8x + 16 = β Β± β x2 x2 β β Isolate the i term Subtract 4 from both sides Square both sides i2 = 1 25 Make equal to zero β + 25 + 25 Add 25 to both sides 8x + 41 = 0 Our Solution x2 β Example 477. The solutions are 5i 3 β 2 and x = 3 + 5i 2 3 5i Β± 2 Write as β²β²oneβ²β² expression equal to x Clear fraction by multiplying by denominator Β± 5i 2x = 3 3 3 β β 2x 5i 3 = 12x + 9 = 5i2 Β± β Isolate the i term Subtract 3 from both sides Square both sides i2 = 1 25 Make equal to zero β 4x2 4x2 β β 12x + 9 = β + 25 + 25 Add 25 to both sides 12x + 34 = 0 Our Solution 4x2 β 350 9.5 Practice - Build Quadratics from Roots From each problem, ο¬nd a quadratic equation with those numbers as its solutions. 1) 2, 5 3) 20, 2 5) 4, 4 7) 0, 0 9) β 11) 3 4 4, 11, 1 4 13) 1 2, 1 3 15) 3 7, 4 17) 19) 21) 23) 25) 27) 29) 1 3, 5 6 6, 1 9 5 1 5 11β 3β 4 i 13β β β Β± Β± Β± Β± Β± 31) 2 33) 1 35) 6 6β 3i i 3β Β± Β± Β± 37) β 6β 1 Β± 2 39) 6 Β± i 2β 8 2) 3, 6 4) 13, 1 6) 0, 9 8) β 10) 3, 2, 5 β 1 β, 5 7 12) 5 8 14) 1 2, 2 3 16) 2, 2 9 18) 5 3, 1 2 β 20) 22) 24) 26) 28) 30) 32) 34β 2 3β 11i 5i 2β 2β 4i i 5β 3 2 9 Β± Β± Β± 36) β 38) 2 Β± 3 5i 40) |
β 2 i 15β Β± 2 351 9.6 Quadratics - Quadratic in Form Objective: Solve equations that are quadratic in form by substitution to create a quadratic equation. We have seen three diο¬erent ways to solve quadratics: factoring, completing the square, and the quadratic formula. A quadratic is any equation of the form 0 = a x2 + bx + c, however, we can use the skills learned to solve quadratics to solve problems with higher (or sometimes lower) powers if the equation is in what is called quadratic form. Quadratic Form: 0 = axm + bxn + c where m = 2n An equation is in quadratic form if one of the exponents on a variable is double the exponent on the same variable somewhere else in the equation. If this is the case we can create a new variable, set it equal to the variable with smallest exponent. When we substitute this into the equation we will have a quadratic equation we can solve. World View Note: Arab mathematicians around the year 1000 were the ο¬rst to use this method! Example 478. β x4 13x2 + 36 = 0 Quadratic form, one exponent, 4, double the other, 2 y = x2 New variable equal to the variable with smaller exponent y2 = x4 y2 13y + 36 = 0 β 9)(y (y 4) = 0 β β 9 = 0 or or y = 4 9 = x2 or 4 = x2 4β = x2β 3, Square both sides Substitute y for x2 and y2 for x4 Solve. We can solve this equation by factoring Set each factor equal to zero Solve each equation Solutions for y, need x. We will use y = x2 equation Substitute values for y Solve using the even root property, simplify roots 2 Our Solutions Β± β = x2β or Β± When we have higher powers of our variable, we could end up with many more solutions. The previous equation had four unique solutions. 352 Example 479. 1 β 2 aβ 1 aβ 6 = 0 Quadratic form, one exponent, 2, is double the other, 1 Make a new variable equal to the variable with lowest exponent β b2 = aβ b2 b 6 = 0 β (b 3)(b + 2) = 0 |
3 = 0 or Raise both sides to 2 = aβ 1 = a 2)β 1 1 3 2 Square both sides 2 and b for aβ Substitute b2 for aβ Solve. We will solve by factoring Set each factor equal to zero Solve each equation Solutions for b, still need a, substitute into b = aβ 1 power Simplify negative exponents 2 b = 3 or b = 1 or 3 = aβ 1 = a or ( Our Solution β + β Just as with regular quadratics, these problems will not always have rational solutions. We also can have irrational or complex solutions to our equations. Example 480. 2x4 + x2 = 6 Make equation equal to zero 6 β 2x4 + x2 Subtract 6 from both sides 6 β 6 = 0 Quadratic form, one exponent, 4, double the other, 2 β y = x2 New variable equal variable with smallest exponent y2 = x4 6 = 0 β (2y 3)(y + 2) = 0 3 = 0 or y + 2 = 0 2 2 Square both sides Solve. We will factor this equation Set each factor equal to zero Solve each equation 2y2 + y β β 2y β + 3 + 3 2 2y = 3 or = x2 or 3 2 β 2 = x2 β 2β = x2β Β± β 6β 2 x = Β± 3 2 Β± r = x2β or i 2β, Β± Our Solution 353 or y = 2 We have y, still need x. Substitute into y = x2 Square root of each side Simplify each root, rationalize denominator When we create a new variable for our substitution, it wonβt always be equal to just another variable. We can make our substitution variable equal to an expression as shown in the next example. Example 481. 3(x 7)2 β β 2(x β 7) + 5 = 0 Quadratic form y = x 7 Deο¬ne new variable Square both sides Substitute values into original Factor Set each factor equal to zero Solve each equation β 3y2 y2 = (x β 7)2 β 2y + 5 = 0 β 5)(y + 1) = 0 (3y 5 = 0 or y + 1 = 0 1 1 1 3y = 5 or 3y y = 3 5 3 7 or = x 5 3 21 3 + β + 7 or |
y = β 1 We have y, we still need x. 1 = x β + 7 7 Substitute into y = x 7 β + 7 Add 7. Use common denominator as needed β x = 26 3, 6 Our Solution Example 482. (x2 β 6x)2 = 7(x2 7(x2 6x) + 12 6x)2 β 7(x2 β β β β 7(x2 (x2 β β 12 Make equation equal zero 6x) 6x) + 12 Move all terms to left β β 6x) + 12 = 0 Quadratic form y = x2 y2 = (x2 y2 β 3)(y (y β 3 = 0 or y β 6x)2 β 7y + 12 = 0 4 6x Make new variable Square both sides Substitute into original equation Solve by factoring Set each factor equal to zero Solve each equation We have y, still need x. Solve each equation, complete the square 6x 6 = 32 = 9 Add 9 to both sides of each equation y = 3 or y = 4 6x or 4 = x3 2 β 3 = x2 β 1 2 Β· 12 = x2 β 6x + 9 or 13 = x2 6x + 9 Factor β 354 Β± 12β = (x 12 = (x 3)2 β 2 3β = x p + 3 Β± 3)2 or 13 = (x β 13β = (x or Β± 3 or β + 3 Β± β β 13β = x p + 3 13β, 3 Β± 3)2 Use even root property 3)2 Simplify roots 3 Add 3 to both sides Our Solution The higher the exponent, the more solution we could have. This is illustrated in the following example, one with six solutions. Example 483. x6 β y β + 1 + 1 9x3 + 8 = 0 Quadratic form, one exponent, 6, double the other, 3 y2 β 1)(y (y β 1 = 0 or y y = x3 New variable equal to variable with lowest exponent y2 = x6 9y + 8 = 0 8 or y = 8 x3 = 1 or x3 = 8 1 8 8 β 8 = 0 Square both sides Substitute y2 for x6 and y for x3 Solve. We will solve by factoring. Set each factor equal to zero Solve each equation Solutions for y, we |
need x. Substitute into y = x3 Set each equation equal to zero 1 Factor each equation, diο¬erence of cubes First equation factored. Set each factor equal to zero First equation is easy to solve x x3 (x β β β β 1 = 0 or x3 β 1)(x2 + x + 1) = 0 β 1 = 0 or x2 + 12 β p2(1) 2)(x2 + 2x + 4) = 0 (x β 2 = 0 or x2 + 2x + 4 = 0 i 3β Β± 2 4(1)(1) = 1 4(1)(4) = 1 i 3β i 3 22 β p2(1) x = 1, 2, 2 β Β± First solution Quadratic formula on second factor Factor the second diο¬erence of cubes Set each factor equal to zero. First equation is easy to solve Our fourth solution Quadratic formula on second factor Our ο¬nal six solutions 355 9.6 Practice - Quadratic in Form Solve each of the following equations. Some equations will have 2) y4 4) y4 6) b4 β β β 8) y4 β 10) x6 14) xβ 9y2 + 20 = 0 29y2 + 100 = 0 10b2 + 9 = 0 40y2 + 144 = 0 β β 2 35x3 + 216 = 0 2y2 = 24 1 xβ 12 = 0 β 20 = 21yβ 1 β 2 β 7x2 + 12 = 0 216 = 19z3 12) y4 1 3 16) 5yβ 3 = 8 18) x4 β 2x2 β 3 = 0 20) x4 + 7x2 + 10 = 0 complex roots. 1) x4 β 5x2 + 4 = 0 3) m4 7m2 8 = 0 β β 50a2 + 49 = 0 5) a4 7) x4 β β 25x2 + 144 = 0 9) m4 β 20m2 + 64 = 0 11) z6 β 13) 6z4 z2 = 12 β 2 3 15) x 17) yβ 35 = 2x β 6 + 7yβ 19) x4 β 21) 2x4 23) x4 β 25) 8x6 27) x8 β 5x2 + 2 = 0 β 9x2 + 8 = 0 9x3 + 1 |
= 0 β 17x4 + 16 = 0 22) 2x4 x2 3 = 0 β β 10x3 + 16 = 0 24) x6 β 26) 8x6 + 7x3 1 = 0 β 4(x 28) (x 1)2 β β 1) = 5 β 30) (x + 1)2 + 6(x + 1) + 9 = 0 32) (m 1)2 β 34) (a + 1)2 + 2(a β 5(m 1) = 14 β 1) = 15 β 1) = 3 36) 2(x 38) (x2 1)2 3)2 β β β β (x β 2(x2 β 3) = 3 40) (x2 + x + 3)2 + 15 = 8(x2 + x + 3) 42) (x2 + x)2 β 8(x2 + x) + 12 = 0 44) (2x2 + 3x)2 = 8(2x2 + 3x) + 9 29) (y + b)2 31) (y + 2)2 33) (x 3)2 β β β β 35) (r 1)2 β β 37) 3(y + 1)2 4(y + b) = 21 6(y + 2) = 16 2(x 8(r 3) = 35 β 1) = 20 β 14(y + 1) = 5 β 39) (3x2 β 2x)2 + 5 = 6(3x2 2x) β 1 5(3x + 1) 3 = 88 2(x2 + 2x) = 3 41) 2(3x + 1) 2 3 43) (x2 + 2x)2 45) (2x2 x)2 β β β β 4(2x2 β x) + 3 = 0 46) (3x2 β 4x)2 = 3(3x2 4x) + 4 β 356 9.7 Quadratics - Rectangles Objective: Solve applications of quadratic equations using rectangles. An application of solving quadratic equations comes from the formula for the area of a rectangle. The area of a rectangle can be calculated by multiplying the width by the length. To solve problems with rectangles we will ο¬rst draw a picture to represent the problem and use the picture to set up our equation. Example 484 |
. The length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions? 40 x + 3 x We do not know the width, x. Length is 4 more, or x + 4, and area is 40. 357 x(x + 3) = 40 Multiply length by width to get area x2 + 3x = 40 Distribute 40 40 Make equation equal zero β β x2 + 3x β 40 = 0 β (x 5)(x + 8) = 0 5 = 0 or x + 8 = 0 8 8 Our x is a width, cannot be negative. Factor Set each factor equal to zero Solve each equation 8 x = 5 or 5) + 3 = 8 5 in by 8in Our Solution Length is x + 3, substitute 5 for x to ο¬nd length The above rectangle problem is very simple as there is only one rectangle involved. When we compare two rectangles, we may have to get a bit more creative. Example 485. If each side of a square is increased by 6, the area is multiplied by 16. Find the side of the original square. x2 x x Square has all sides the same length Area is found by multiplying length by width 16x2 x + 6 Each side is increased by 6, x + 6 Area is 16 times original area (x + 6)(x + 6) = 16x2 Multiply length by width to get area x2 + 12x + 36 = 16x2 FOIL 16x2 β β 16x2 Make equation equal zero β 15x2 + 12x + 36 = 0 Divide each term by 15x2 12)2 12x β 4(15)( 36 = 0 36) β Solve using the quadratic formula 1, changes the signs β Evaluate 12 Β± x = ( p β β β 2(15) 16 x = Β± 2304β 30 16 48 Β± 30 60 30 x = Canβ²t have a negative solution, we will only add x = = 2 Our x is the original square 358 2 Our Solution Example 486. The length of a rectangle is 4 ft greater than the width. If each dimension is increased by 3, the new area will be 33 square feet larger. Find the dimensions of the original rectangle. x(x + 4) x We donβ²t know width, x, length is 4 more, x + 4 x + 4 Area is found by multiplying length by width x |
(x + 4) + 33 x + 3 Increase each side by 3. width becomes x + 3, length x + 3)(x + 7) = x(x + 4) + 33 x2 + 10x + 21 = x2 + 4x + 33 x2 x2 β β Area is 33 more than original, x(x + 4) + 33 Set up equation, length times width is area Subtract x2 from both sides β β 10x + 21 = 4x + 33 Move variables to one side 4x Subtract 4x from each side Subtract 21 from both sides 4x 6x + 21 = 33 21 21 6x = 12 Divide both sides by 6 6 β β 6 x = 2 (2) + 4 = 6 2 ft by 6ft Our Solution x is the width of the original x + 4 is the length. Substitute 2 to ο¬nd From one rectangle we can ο¬nd two equations. Perimeter is found by adding all the sides of a polygon together. A rectangle has two widths and two lengths, both the same size. So we can use the equation P = 2l + 2w (twice the length plus twice the width). Example 487. The area of a rectangle is 168 cm2. The perimeter of the same rectangle is 52 cm. What are the dimensions of the rectangle? x We donβ²t know anything about length or width y xy = 168 Use two variables, x and y Length times width gives the area. 2x + 2y = 52 Also use perimeter formula. 2x β 2y = 2x + 52 Divide each term by 2 Solve by substitution, isolate y 2x β β 359 2 2 2 x + 26 x( Substitute into area equation β y = x + 26) = 168 Distribute β x2 + 26x = 168 Divide each term by β x2 26 168 26x = β 2 β = 132 = 169 1 2 Β· x2 Solve by completing the square. 1, changing all the signs β Find number to complete the square: 1 2 Β· b 2 β 26x + 324 = 1 Add 169 to both sides (x Factor x Square root both sides 13)2 = 1 β 1 13 = β + 13 + 13 x = 13 Β± Β± 1 Evaluate y = y = β β x = 14 or 12 Two options for ο¬rst side. (14) + 26 = 12 (12) + 26 |
= 14 Substitute 14 into y = β Substitute 12 into y = β Both are the same rectangle, variables switched! x + 26 x + 26 12 cm by 14cm Our Solution World View Note: Indian mathematical records from the 9th century demonstrate that their civilization had worked extensivly in geometry creating religious alters of various shapes including rectangles. Another type of rectangle problem is what we will call a βframe problemβ. The idea behind a frame problem is that a rectangle, such as a photograph, is centered inside another rectangle, such as a frame. In these cases it will be important to rememember that the frame extends on all sides of the rectangle. This is shown in the following example. Example 488. An 8 in by 12 in picture has a frame of uniform width around it. The area of the frame is equal to the area of the picture. What is the width of the frame? 8 12 12 + 2x Draw picture, picture if 8 by 10 If frame has width x, on both sides, we add 2x 8 + 2x 8 2 Β· 12 = 96 Area of the picture, length times width Β· 96 = 192 Frame is the same as the picture. Total area is double this. (12 + 2x)(8 + 2x) = 192 Area of everything, length times width 96 + 24x + 16x + 4x2 = 192 FOIL 4x2 + 40x + 96 = 192 Combine like terms 192 192 Make equation equal to zero by subtracting 192 β 4x2 + 40x β 96 = 0 β Factor out GCF of 4 360 x β β 4(x2 + 10x 4(x 24) = 0 2)(x + 12) = 0 2 = 0 or x + 12 = 0 12 12 Canβ²t have negative frame width. Factor trinomial Set each factor equal to zero Solve each equation 12 x = 2 or β β + 2 + 2 β β 2 inches Our Solution Example 489. A farmer has a ο¬eld that is 400 rods by 200 rods. He is mowing the ο¬eld in a spiral pattern, starting from the outside and working in towards the center. After an hour of work, 72% of the ο¬eld is left uncut. What is the size of the ring cut around the outside? 200 β 2x 200 Draw picture, outside is 200 by 400 If frame has width x on both sides, subtract 2x from each |
side to get center 400 2x β 400 400 200 = 80000 Area of entire ο¬eld, length times width (0.72) = 57600 Area of center, multiply by 28% as decimal Β· 2x) = 57600 Area of center, length times width 80000 Β· 2x)(200 β (400 β 800x β 4x2 β β 80000 400x + 4x2 = 57600 FOIL 1200x + 80000 = 57600 Combine like terms 57600 57600 Make equation equal zero β β 4x2 β 4(x2 4(x 1200x + 22400 = 0 300x + 5600) = 0 280)(x 20) = 0 β 280 = 0 or x 20 = 0 β + 20 + 20 β β β + 280 + 280 x Factor out GCF of 4 Factor trinomial Set each factor equal to zero Solve each equation x = 280 or 20 The ο¬eld is only 200 rods wide, Canβ²t cut 280 oο¬ two sides! 20 rods Our Solution For each of the frame problems above we could have also completed the square or use the quadratic formula to solve the trinomials. Remember that completing the square or the quadratic formula always will work when solving, however, factoring only works if we can factor the trinomial. 361 9.7 Practice - Rectangles 1) In a landscape plan, a rectangular ο¬owerbed is designed to be 4 meters longer than it is wide. If 60 square meters are needed for the plants in the bed, what should the dimensions of the rectangular bed be? 2) If the side of a square is increased by 5 the area is multiplied by 4. Find the side of the original square. 3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square yards. Find the dimensions of the lot. 4) The length of a room is 8 ft greater than it is width. If each dimension is increased by 2 ft, the area will be increased by 60 sq. ft. Find the dimensions of the rooms. 5) The length of a rectangular lot is 4 rods greater than its width, and its area is 60 square rods. Find the dimensions of the lot. 6) The length of a rectangle is 15 ft greater than its width. If each dimension is decreased by 2 ft, the area will be decreased by 106 ft2. Find the dimensions. |
7) A rectangular piece of paper is twice as long as a square piece and 3 inches wider. The area of the rectangular piece is 108 in2. Find the dimensions of the square piece. 8) A room is one yard longer than it is wide. At 75c per sq. yd. a covering for the ο¬oor costs S31.50. Find the dimensions of the ο¬oor. 9) The area of a rectangle is 48 ft2 and its perimeter is 32 ft. Find its length and width. 10) The dimensions of a picture inside a frame of uniform width are 12 by 16 inches. If the whole area (picture and frame) is 288 in2, what is the width of the frame? 11) A mirror 14 inches by 15 inches has a frame of uniform width. If the area of the frame equals that of the mirror, what is the width of the frame. 12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when mowing the grass to have cut half of it. 13) A grass plot 9 yards long and 6 yards wide has a path of uniform width around it. If the area of the path is equal to the area of the plot, determine the width of the path. 14) A landscape architect is designing a rectangular ο¬owerbed to be border with 28 plants that are placed 1 meter apart. He needs an inner rectangular space in the center for plants that must be 1 meter from the border of the bed and 362 that require 24 square meters for planting. What should the overall dimensions of the ο¬owerbed be? 15) A page is to have a margin of 1 inch, and is to contain 35 in2 of painting. How large must the page be if the length is to exceed the width by 2 inches? 16) A picture 10 inches long by 8 inches wide has a frame whose area is one half the area of the picture. What are the outside dimensions of the frame? 17) A rectangular wheat ο¬eld is 80 rods long by 60 rods wide. A strip of uniform width is cut around the ο¬eld, so that half the grain is left standing in the form of a rectangular plot. How wide is the strip that is cut? 18) A picture 8 inches by 12 inches is placed in a frame of uniform width. If the area of the frame equals the area of the picture ο¬nd the width of the frame. 19) |
A rectangular ο¬eld 225 ft by 120 ft has a ring of uniform width cut around the outside edge. The ring leaves 65% of the ο¬eld uncut in the center. What is the width of the ring? 20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft. He starts cutting around the outside boundary spiraling around towards the center. By noon he has cut 60% of the lawn. What is the width of the ring that he has cut? 21) A frame is 15 in by 25 in and is of uniform width. The inside of the frame leaves 75% of the total area available for the picture. What is the width of the frame? 22) A farmer has a ο¬eld 180 ft by 240 ft. He wants to increase the area of the ο¬eld by 50% by cultivating a band of uniform width around the outside. How wide a band should he cultivate? 23) The farmer in the previous problem has a neighber who has a ο¬eld 325 ft by 420 ft. His neighbor wants to increase the size of his ο¬eld by 20% by cultivating a band of uniform width around the outside of his lot. How wide a band should his neighbor cultivate? 24) A third farmer has a ο¬eld that is 500 ft by 550 ft. He wants to increase his ο¬eld by 20%. How wide a ring should he cultivate around the outside of his ο¬eld? 25) Donna has a garden that is 30 ft by 36 ft. She wants to increase the size of the garden by 40%. How wide a ring around the outside should she cultivate? 26) A picture is 12 in by 25 in and is surrounded by a frame of uniform width. The area of the frame is 30% of the area of the picture. How wide is the frame? 363 9.8 Quadratics - Teamwork Objective: Solve teamwork problems by creating a rational equation to model the problem. If it takes one person 4 hours to paint a room and another person 12 hours to paint the same room, working together they could paint the room even quicker, it turns out they would paint the room in 3 hours together. This can be reasoned by the following logic, if the ο¬rst person paints the room in 4 hours, she paints 1 4 of the room each hour. If the second person takes 12 hours to paint the room |
, he paints 1 4 + 1 12 of the room each hour. So together, each hour they paint 1 12 of the room. Using a common denominator of 12 gives: 3 12 = 1 3. This means each hour, working together they complete 1 3 is completed each hour, it follows that it will take 3 hours to complete the entire room. 12 + 1 12 = 4 3 of the room. If 1 This pattern is used to solve teamwork problems. If the ο¬rst person does a job in A, a second person does a job in B, and together they can do a job in T (total). We can use the team work equation. Teamwork Equation: 1 A + 1 B = 1 T Often these problems will involve fractions. Rather than thinking of the ο¬rst fraction as 1 A, it may be better to think of it as the reciprocal of Aβs time. World View Note: When the Egyptians, who were the ο¬rst to work with fractions, wrote fractions, they were all unit fractions (numerator of one). They only used these type of fractions for about 2000 years! Some believe that this cumbersome style of using fractions was used for so long out of tradition, others believe the Egyptians had a way of thinking about and working with fractions that has been completely lost in history. Example 490. Adam can clean a room in 3 hours. If his sister Maria helps, they can clean it in 2 2 5 hours. How long will it take Maria to do the job alone? 2 2 5 = Adan: 3, Maria: x, Total: 12 5 5 12 Together time, 2 2 5, needs to be converted to fraction Clearly state times for each and total, using x for Maria 364 1 3 + 1 x = 5 12 Using reciprocals, add the individual times gives total 1(12x) 3 + 1(12x) x = 5(12x) 12 Multiply each term by LCD of 12x 4x + 12 = 5x Reduce each fraction 4x 4x Move variables to one side, subtracting 4x β β 12 = x Our solution for x It takes Maria 12 hours Our Solution Somtimes we only know how two peopleβs times are related to eachother as in the next example. Example 491. Mike takes twice as long as Rachel to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the |
project alone? Mike: 2x, Rachel: x, Total: 10 Clearly deο¬ne variables. If Rachel is x, Mike is 2x 1 2x + 1 x = 1 10 Using reciprocals, add individal times equaling total 1(10x) 2x + 1(10x) x = 1(10x) 10 Multiply each term by LCD, 10x 5 + 10 = x Combine like terms 15 = x We have our x, we said x was Rachelβ²s time 2(15) = 30 Mike is double Rachel, this gives Mikeβ²s time. Mike: 30 hr, Rachel: 15hr Our Solution With problems such as these we will often end up with a quadratic to solve. Example 492. Brittney can build a large shed in 10 days less than Cosmo can. If they built it together it would take them 12 days. How long would it take each of them working alone? Britney: x β 10, Cosmo: x, Total: 12 1 12 1 x 10 = + x 1 β 365 If Cosmo is x, Britney is x 10 β Using reciprocals, make equation 1(12x(x x β β 10 10)) + 1(12x(x x β 10)) = 1(12x(x 12 β 10)) Multiply by LCD: 12x(x 10) β 12x + 12(x β 12x + 12x β 24x β 24x + 120 10) = x(x 120 = x2 120 = x2 β β β β 0 = x2 0 = (x 30 = 0 or x x β + 30 + 30 β β β 10) Reduce fraction 10x Distribute 10x Combine like terms 24x + 120 Move all terms to one side 34x + 120 30)(x 4 Factor Set each factor equal to zero Solve each equation x = 30 or x = 4 This, x, was deο¬ned as Cosmo. 30 10 = 20 or 4 10 = 6 Find Britney, canβ²t have negative time β β β Britney: 20 days, Cosmo: 30 days Our Solution In the previous example, when solving, one of the possible times ended up negative. We canβt have a negative amount of time to build a shed, so this possibility is ignored for this problem. Also, as we were solving, we had to factor x2 34x + 120. This may have been |
diο¬cult to factor. We could have also chosen to complete the square or use the quadratic formula to ο¬nd our solutions. β It is important that units match as we solve problems. This means we may have to convert minutes into hours to match the other units given in the problem. Example 493. An electrician can complete a job in one hour less than his apprentice. Together they do the job in 1 hour and 12 minutes. How long would it take each of them working alone? 1 hr 12 min = 1 12 60 hr Change 1 hour 12 minutes to mixed number 1 12 60 = 1 1 5 = Electrician: x β 1, Apprentice: x, Total Reduce and convert to fraction Clearly deο¬ne variables Using reciprocals, make equation 1(6x(x x β β 1 1)) + 1(6x(x x β 1)) = 5(6x(x 6 β 1) Multiply each term by LCD 6x(x 1) β 366 6x + 6(x 1) = 5x(x 6 = 5x2 6 = 5x2 β β β β 6x + 6x 12x 12x + 6 0 = 5x2 β β β 1) Reduce each fraction 5x Distribute 5x Combine like terms 5x 0 = (5x 2 = 0 or x β β β 2)(x Factor Set each factor equal to zero Solve each equation 12x + 6 Move all terms to one side of equation 17x + 6 3 5x = 2 or x = 3 5 β + 2 + 2 or x = 3 Subtract 1 from each to ο¬nd electrician or 3 β Electrician: 2 hr, Apprentice: 3 hours Our Solution 1 = 2 Ignore negative. Very similar to a teamwork problem is when the two involved parts are working against each other. A common example of this is a sink that is ο¬lled by a pipe and emptied by a drain. If they are working against eachother we need to make one of the values negative to show they oppose eachother. This is shown in the next example.. Example 494. A sink can be ο¬lled by a pipe in 5 minutes but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to ο¬ll the sink? Sink: 5, Drain: 7, Total: x Deο¬ |
ne variables, drain is negative 1 5 β 1 7 = 1 x 1(35x) 5 β 1(35x) 7 = 1(35x) x Using reciprocals to make equation, Subtract because they are opposite Multiply each term by LCD: 35x 7x β 5x = 35 Reduce fractions 2x = 35 Combine like terms 2 x = 17.5 Our answer for x Divide each term by 2 2 17.5 min or 17 min 30 sec Our Solution 367 9.8 Practice - Teamwork 1) Bills father can paint a room in two hours less than Bill can paint it. Working together they can complete the job in two hours and 24 minutes. How much time would each require working alone? 2) Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to ο¬ll a pool. When both pipes are open, the pool is ο¬lled in three hours and forty-ο¬ve minutes. If only the larger pipe is open, how many hours are required to ο¬ll the pool? 3) Jack can wash and wax the family car in one hour less than Bob can. The two working together can complete the job in 1 1 each require if they worked alone? 5 hours. How much time would 4) If A can do a piece of work alone in 6 days and B can do it alone in 4 days, how long will it take the two working together to complete the job? 5) Working alone it takes John 8 hours longer than Carlos to do a job. Working together they can do the job in 3 hours. How long will it take each to do the job working alone? 6) A can do a piece of work in 3 days, B in 4 days, and C in 5 days each working alone. How long will it take them to do it working together? 7) A can do a piece of work in 4 days and B can do it in half the time. How long will it take them to do the work together? 8) A cistern can be ο¬lled by one pipe in 20 minutes and by another in 30 minutes. How long will it take both pipes together to ο¬ll the tank? 9) If A can do a piece of work in 24 days and A and B together can do it in 6 days, how long would it take B to do the work alone? 10) A carpenter and his assistant can do a piece of work in 3 |
3 4 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone? 11) If Sam can do a certain job in 3 days, while it takes Fred 6 days to do the same job, how long will it take them, working together, to complete the job? 12) Tim can ο¬nish a certain job in 10 hours. It take his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job? 13) Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the faster person, working alone, to do the job? 14) If two people working together can do a job in 3 hours, how long will it take the slower person to do the same job if one of them is 3 times as fast as the other? 15) A water tank can be ο¬lled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long will it take to ο¬ll the tank if both pipes are open? 368 16) A sink can be ο¬lled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink? 17) It takes 10 hours to ο¬ll a pool with the inlet pipe. It can be emptied in 15 hrs with the outlet pipe. If the pool is half full to begin with, how long will it take to ο¬ll it from there if both pipes are open? 18) A sink is 1 4 full when both the faucet and the drain are opened. The faucet alone can ο¬ll the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to ο¬ll the remaining 3 4 of the sink? 19) A sink has two faucets, one for hot water and one for cold water. The sink can be ο¬lled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is ο¬lled in 2.1 minutes. How long does it take to οΏ½ |
οΏ½οΏ½ll the sink with just the hot-water faucet open? 20) A water tank is being ο¬lled by two inlet pipes. Pipe A can ο¬ll the tank in 4 1 2 hrs, while both pipes together can ο¬ll the tank in 2 hours. How long does it take to ο¬ll the tank using only pipe B? 21) A tank can be emptied by any one of three caps. The ο¬rst can empty the tank in 20 minutes while the second takes 32 minutes. If all three working together could empty the tank in 8 8 to empty the tank? 59 minutes, how long would the third take 22) One pipe can ο¬ll a cistern in 1 1 3 hrs. Three pipes working together ο¬ll the cistern in 42 minutes. How long would it take the third pipe alone to ο¬ll the tank? 2 hours while a second pipe can ο¬ll it in 2 1 23) Sam takes 6 hours longer than Susan to wax a ο¬oor. Working together they can wax the ο¬oor in 4 hours. How long will it take each of them working alone to wax the ο¬oor? 24) It takes Robert 9 hours longer than Paul to rapair a transmission. If it takes 5 hours to do the job if they work together, how long will it take each them 2 2 of them working alone? 25) It takes Sally 10 1 2 minutes longer than Patricia to clean up their dorm room. If they work together they can clean it in 5 minutes. How long will it take each of them if they work alone? 26) A takes 7 1 2 minutes longer than B to do a job. Working together they can do the job in 9 minutes. How long does it take each working alone? 27) Secretary A takes 6 minutes longer than Secretary B to type 10 pages of manuscript. If they divide the job and work together it will take them 8 3 4 minutes to type 10 pages. How long will it take each working alone to type the 10 pages? 28) It takes John 24 minutes longer than Sally to mow the lawn. If they work together they can mow the lawn in 9 minutes. How long will it take each to mow the lawn if they work alone? 369 9.9 Quadratics - Simultaneous Products Objective: Solve simultaneous product equations using substitution to create a rational equation. When solving a system of equations where |
the variables are multiplied together we can use the same idea of substitution that we used with linear equations. When we do so we may end up with a quadratic equation to solve. When we used substitution we solved for a variable and substitute this expression into the other equation. If we have two products we will choose a variable to solve for ο¬rst and divide both sides of the equations by that variable or the factor containing the variable. This will create a situation where substitution can easily be done. Example 495. xy = 48 (x + 3)(y 2) = 54 β To solve for x, divide ο¬rst equation by x, second by x + 3 y = 48 x and y 2 = β 54 x + 3 Substitute 48 x for y in the second equation 48 x β 2 = 54 x + 3 Multiply each term by LCD: x(x + 3) 48x(x + 3) x β 2x(x + 3) = 54x(x + 3) x + 3 Reduce each fraction 48(x + 3) 48x + 144 2x(x + 3) = 54x Distribute 2x2 6x = 54x Combine like terms β β β 2x2 + 42x + 144 = 54x Make equation equal zero β 2 β 54x 54x Subtract 54x from both sides 12x + 144 = 0 Divide each term by GCF of β β 2x2 β β x2 + 6x x β β + 6 + 6 12 x = 6 or x = Factor Set each factor equal to zero Solve each equation 72 = 0 β 6)(x + 12) = 0 (x 6 = 0 or x + 12 = 0 12 β 12 β 12y = 48 12 12 4 Our solutions for y, 6y = 48 or β 6 β β y = 8 or y = β 4) Our Solutions as ordered pairs (6, 8) or ( Substitute each solution into xy = 48 Solve each equation 12, β 6 β β 370 These simultaneous product equations will also solve by the exact same pattern. We pick a variable to solve for, divide each side by that variable, or factor containing the variable. This will allow us to use substitution to create a rational expression we can use to solve. Quite often these problems will have two solutions. Example 496. xy = 35 (x + 6)(y β β 2) = 5 To solve for x, divide the οΏ½ |
οΏ½οΏ½rst equation by x, second by x + 6 35 y = β x and y 2 = β 5 x + 6 35 Substitute β x for y in the second equation β 35 x β 2 = 5 x + 6 Multiply each term by LCD: x(x + 6) β 35x(x + 6) x β 2x(x + 6) = 5x(x + 6) x + 6 Reduce fractions β β 35(x + 6) 35x 210 2x2 β β 12x = 5x Combine like terms 210 = 5x Make equation equal zero 2x(x + 6) = 5x Distribute 2x2 47x 5x 52x β β 5x β 210 = 0 Divide each term by 2 β β β β β β Factor Set each factor equal to zero Solve each equation β 2x2 β x2 + 26x + 105 = 0 (x + 5)(x + 21) = 0 x + 5 = 0 or x + 21 = 0 21 21 35 21 5 3 β β β β y = 7 or y = 21 β 5 or x = 21y = 21 5 β x = β 35 or 5 β β β β β 5 5, 7) or ( β 21, 5 3 β β 5y = 5 β Substitute each solution into xy = Solve each equation 35 β Our solutions for y Our Solutions as ordered pairs The processes used here will be used as we solve applications of quadratics including distance problems and revenue problems. These will be covered in another section. World View Note: William Horner, a British mathematician from the late 18th century/early 19th century is credited with a method for solving simultaneous equations, however, Chinese mathematician Chu Shih-chieh in 1303 solved these equations with exponents as high as 14! 371 9.9 Practice - Simultaneous Product Solve. 1) 3) xy = 72 (x + 2)(y β xy = 150 4) = 128 6)(y + 1) = 64 (x β 5) xy = 45 (x + 2)(y + 1) = 70 7) xy = 90 5)(y + 1) = 120 (x β 9) 11) xy = 12 (x + 1)(y β xy = 45 4) = 16 5)(y + 3) = 160 (x β 2) 4) 6) 8) 10) (x xy = 180 1)( |
y β β xy = 120 1 2) = 205 (x + 2)(y 3)=120 β xy = 65 (x 8)(y + 2) = 35 β xy = 48 (x 6)(y + 3) = 60 β xy = 60 (x + 5)(y + 3) = 150 12) xy = 80 5)(y + 5) = 45 (x β 372 9.10 Quadratics - Revenue and Distance Objective: Solve revenue and distance applications of quadratic equations. A common application of quadratics comes from revenue and distance problems. Both are set up almost identical to each other so they are both included together. Once they are set up, we will solve them in exactly the same way we solved the simultaneous product equations. Revenue problems are problems where a person buys a certain number of items for a certain price per item. If we multiply the number of items by the price per item we will get the total paid. To help us organize our information we will use the following table for revenue problems Number Price Total First Second The price column will be used for the individual prices, the total column is used for the total paid, which is calculated by multiplying the number by the price. Once we have the table ο¬lled out we will have our equations which we can solve. This is shown in the following examples. Example 497. A man buys several ο¬sh for S56. After three ο¬sh die, he decides to sell the rest at a proο¬t of S5 per ο¬sh. His total proο¬t was S4. How many ο¬sh did he buy to begin with? Buy Sell Buy Sell Number Price Total n p 56 Using our table, we donβ²t know the number he bought, or at what price, so we use varibles n and p. Total price was S56. Number Price Total n β 3 n p p + 5 56 60 When he sold, he sold 3 less (n more (p + 5). Total proο¬t was S4, combined with S56 spent is S60 3), for S5 β np = 56 3)(p + 5) = 60 Find equatinos by multiplying number by price These are a simultaneous product (n β p = 56 n and p + 5 = 60 β n 3 Solving for number, divide by n or (n 3) β 373 56 n + 5 = 60 β n |
3 Substitute 56 n for p in second equation 56n(n n β 3) + 5n(n 3) = β 3) 60n(n n β β 3 Multiply each term by LCD: n(n 3) β 56(n 56n β 3) + 5n(n β 168 + 5n2 5n2 + 41n 60n β β 3) = 60n β 15n = 60n 168 = 60n Move all terms to one side Reduce fractions Combine like terms β 5n2 β 19)2 19n β 4(5)( β 2(5) β 60n 168 = 0 168) β Solve with quadratic formula Simplify 19 Β± n = ( p β n = 19 3721β 10 Β± 19 = 61 Β± 10 We donβ²t want negative solutions, only do + n = 80 10 = 8 This is our n 8 ο¬sh Our Solution Example 498. A group of students together bought a couch for their dorm that cost S96. However, 2 students failed to pay their share, so each student had to pay S4 more. How many students were in the original group? Number Price Total n p 96 Deal Paid Number Price Total Deal Paid n β 2 n p p + 4 96 96 S96 was paid, but we donβ²t know the number or the price agreed upon by each student. There were 2 less that actually paid (n β and they had to pay S4 more (p + 4). The total here is still S96. 2) np = 96 2)(p + 4) = 96 Equations are product of number and price This is a simultaneous product (n β p = 96 n and p + 4 = 96 n + 4 = 96 β 96 β n n 2 2 Solving for number, divide by n and n 2 β Substitute 96 n for p in the second equation 374 96n(n n β 2) + 4n(n 2) = β 2) 96n(n n β β 2 Multiply each term by LCD: n(n 2) β 96(n 96n 2) + 4n(n β 192 + 4n2 β 4n2 + 88n 96n β 4n2 β β β 2) = 96n 8n = 96n 192 = 96n 96n 8n 192 = 0 + 192 + 192 8n = 192 4 4 β β β β 4n2 4 Reduce fractions Distribute Combine |
like terms Set equation equal to zero Solve by completing the square, Separate variables and constant Divide each term by a or 4 2 1 2 Β· n2 2n = 48 Complete the square: b β 2 2 n2 Β· = 12 = 1 Add to both sides of equation 1 2 2n + 1 = 49 β 1)2 = 49 ( We donβ²t want a negative solution n = 1 + 7 = 8 Factor Square root of both sides Add 1 to both sides Β± Β± 8 students Our Solution The above examples were solved by the quadratic formula and completing the square. For either of these we could have used either method or even factoring. Remember we have several options for solving quadratics. Use the one that seems easiest for the problem. Distance problems work with the same ideas that the revenue problems work. The only diο¬erence is the variables are r and t (for rate and time), instead of n and p (for number and price). We already know that distance is calculated by multiplying rate by time. So for our distance problems our table becomes the following: rate time distance First Second Using this table and the exact same patterns as the revenue problems is shown in the following example. Example 499. 375 Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference? There Back There Back r rate time distance r t 120 We do not know rate, r, or time, t he traveled on the way to the conference. But we do know the distance was 120 miles. time distance rate r t 10 t + 2 β 120 120 Coming back he drove 10 mph slower (r 10) and took 2 hours longer (t + 2). The distance was still 120 miles. β rt = 120 Equations are product of rate and time 10)(t + 2) = 120 We have simultaneous product equations (r β t = 120 r and t + 2 = 120 r + 2 = 120 r r 10 β 120 10 β Solving for rate, divide by r and r 10 β Substitute 120 r for t in the second equation 120r(r r β 10) + 2r(r 10) = β 120r(r r β 10) β 10 Multiply each term by LCD: r(r 10) β Reduce each fraction Distribute Combine like terms 120r Make equation |
equal to zero Divide each term by 2 Factor Set each factor equal to zero Solve each equation 120(r 120r β β 20r = 120r β 20r = 120r β 1200 = 120r 10) + 2r2 β 1200 + 2r2 β 2r2 + 100r 120r β 2r2 20r 1200 = 0 β β r2 10r 600 = 0 β β 30)(r + 20) = 0 30 = 0 and r + 20 = 0 20 β 20 β 30 mph 20 r = 30 and r = β + 30 + 30 (r β β r Canβ²t have a negative rate Our Solution World View Note: The worldβs fastest man (at the time of printing) is Jamaican Usain Bolt who set the record of running 100 m in 9.58 seconds on August 16, 2009 in Berlin. That is a speed of over 23 miles per hour! Another type of simultaneous product distance problem is where a boat is traveling in a river with the current or against the current (or an airplane ο¬ying with the wind or against the wind). If a boat is traveling downstream, the current will push it or increase the rate by the speed of the current. If a boat is traveling 376 upstream, the current will pull against it or decrease the rate by the speed of the current. This is demonstrated in the following example. Example 500. A man rows down stream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current ο¬ows at 2 miles per hour, how fast would the man row in still water? 8 rate time distance down up rate down r + 2 up 2 8 r β t 8 t β 30 30 time distance t t β 30 30 Write total time above time column We know the distance up and down is 30. Put t for time downstream. Subtracting 8 t becomes time upstream β Downstream the current of 2 mph pushes the boat (r + 2) and upstream the current pulls the boat (r 2) β (r + 2)t = 30 Multiply rate by time to get equations t) = 30 We have a simultaneous product 2)(8 β (r β t = 30 r + 2 and 8 t = β 30 r + 2 = 8 β 30 β 30 β 2 2 r r Solving for rate, divide by r + 2 or r 2 β Substitute 30 r + 2 for t in second equation 8(r + 2 |
)(r 2) β β 30(r + 2)(r r + 2 2) = β 30(r + 2)(r r 2 β 2) β Multiply each term by LCD: (r + 2)(r 2) β 8(r + 2)(r β 8r2 β 2) β 32 8r2 β 30(r 2) = 30(r + 2) Reduce fractions 30r + 60 = 30r + 60 Multiply and distribute 30r + 28 = 30r + 60 Make equation equal zero 30r 30r β β β 60 60 β β β 8r2 60r 32 = 0 β β 2r2 15r 8 = 0 β β (2r + 1)(r 8) = 0 β 2r + 1 = 0 or or r = 8 β 2 1 2 1 1 β 2r = 2 or r = 8 r = β β 8 mph Divide each term by 4 Factor Set each factor equal to zero Solve each equation Canβ²t have a negative rate Our Solution 377 9.10 Practice - Revenue and Distance 1) A merchant bought some pieces of silk for S900. Had he bought 3 pieces more for the same money, he would have paid S15 less for each piece. Find the number of pieces purchased. 2) A number of men subscribed a certain amount to make up a deο¬cit of S100 but 5 men failed to pay and thus increased the share of the others by S1 each. Find the amount that each man paid. 3) A merchant bought a number of barrels of apples for S120. He kept two barrels and sold the remainder at a proο¬t of S2 per barrel making a total proο¬t of S34. How many barrels did he originally buy? 4) A dealer bought a number of sheep for S440. After 5 had died he sold the remainder at a proο¬t of S2 each making a proο¬t of S60 for the sheep. How many sheep did he originally purchase? 5) A man bought a number of articles at equal cost for S500. He sold all but two for S540 at a proο¬t of S5 for each item. How many articles did he buy? 6) A clothier bought a lot of suits for S750. He sold all but 3 of them for S864 making a proο¬t of S7 on each suit sold. How many suits did he buy |
? 7) A group of boys bought a boat for S450. Five boys failed to pay their share, hence each remaining boys were compelled to pay S4.50 more. How many boys were in the original group and how much had each agreed to pay? 8) The total expenses of a camping party were S72. If there had been 3 fewer persons in the party, it would have cost each person S2 more than it did. How many people were in the party and how much did it cost each one? 9) A factory tests the road performance of new model cars by driving them at two diο¬erent rates of speed for at least 100 kilometers at each rate. The speed rates range from 50 to 70 km/hr in the lower range and from 70 to 90 km/hr in the higher range. A driver plans to test a car on an available speedway by driving it for 120 kilometers at a speed in the lower range and then driving 120 kilometers at a rate that is 20 km/hr faster. At what rates should he drive if he plans to complete the test in 3 1 2 hours? 10) A train traveled 240 kilometers at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hour. What was the rate of each engine? 11) The rate of the current in a stream is 3 km/hr. A man rowed upstream for 3 kilometers and then returned. The round trip required 1 hour and 20 minutes. How fast was he rowing? 378 12) A pilot ο¬ying at a constant rate against a headwind of 50 km/hr ο¬ew for 750 kilometers, then reversed direction and returned to his starting point. He completed the round trip in 8 hours. What was the speed of the plane? 13) Two drivers are testing the same model car at speeds that diο¬er by 20 km/hr. The one driving at the slower rate drives 70 kilometers down a speedway and returns by the same route. The one driving at the faster rate drives 76 kilometers down the speedway and returns by the same route. Both drivers leave at the same time, and the faster car returns 1 slower car. At what rates were the cars driven? 2 hour earlier than the 14) An athlete plans to row upstream a distance of 2 kilometers and then return to his starting point in a total time of 2 hours and 20 minutes. If |
the rate of the current is 2 km/hr, how fast should he row? 15) An automobile goes to a place 72 miles away and then returns, the round trip occupying 9 hours. His speed in returning is 12 miles per hour faster than his speed in going. Find the rate of speed in both going and returning. 16) An automobile made a trip of 120 miles and then returned, the round trip occupying 7 hours. Returning, the rate was increased 10 miles an hour. Find the rate of each. 17) The rate of a stream is 3 miles an hour. If a crew rows downstream for a distance of 8 miles and then back again, the round trip occupying 5 hours, what is the rate of the crew in still water? 18) The railroad distance between two towns is 240 miles. If the speed of a train were increased 4 miles an hour, the trip would take 40 minutes less. What is the usual rate of the train? 19) By going 15 miles per hour faster, a train would have required 1 hour less to travel 180 miles. How fast did it travel? 20) Mr. Jones visits his grandmother who lives 100 miles away on a regular basis. Recently a new freeway has opend up and, although the freeway route is 120 miles, he can drive 20 mph faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones rate on both the old route and on the freeway? 21) If a train had traveled 5 miles an hour faster, it would have needed 1 1 2 hours less time to travel 150 miles. Find the rate of the train. 22) A traveler having 18 miles to go, calculates that his usual rate would make him one-half hour late for an appointment; he ο¬nds that in order to arrive on time he must travel at a rate one-half mile an hour faster. What is his usual rate? 379 9.11 Quadratics - Graphs of Quadratics Objective: Graph quadratic equations using the vertex, x-intercepts, and y-intercept. Just as we drew pictures of the solutions for lines or linear equations, we can draw a picture of solution to quadratics as well. One way we can do that is to make a table of values. Example 501. y = x2 β 4x + 3 Make a table of values x y 0 1 2 3 4 We will test 5 values to get an idea of shape y = (0)2 + 4( |
0) + 3 = 0 y = (1)2 4(1) + 3 = 1 β y = (2)2 4(23)2 4(34)2 4(4) + 3 = 16 β β 0 + 3 = 3 Plug 0 in for x and evaluate β 4 + 3 = 0 Plug 1 in for x and evaluate β 1 Plug 2 in for x and evaluate 8 + 3 = 12 + 3 = 0 Plug 3 in for x and evaluate 16 + 3 = 3 Plug 4 in for x and evaluate β Our completed table. Plot points on graph Plot the points (0, 3), (1, 0), (2, (3, 0), and (4, 3). β 1), Connect curve. the dots with a smooth Our Solution When we have x2 in our equations, the graph will no longer be a straight line. Quadratics have a graph that looks like a U shape that is called a parabola. World View Note: The ο¬rst major female mathematician was Hypatia of Egypt who was born around 370 AD. She studied conic sections. The parabola is one type of conic section. 380 The above method to graph a parabola works for any equation, however, it can be very tedious to ο¬nd all the correct points to get the correct bend and shape. For this reason we identify several key points on a graph and in the equation to help us graph parabolas more eο¬ciently. These key points are described below. Point A: y-intercept: Where the graph crosses the vertical y-axis. Points B and C: x-intercepts: Where the graph crosses the horizontal x-axis Point D: Vertex: The point where the graph curves and changes directions. B C A D We will use the following method to ο¬nd each of the points on our parabola. To graph the equation y = ax2 + b x + c, ο¬nd the following points 1. y-intercept: Found by making x = 0, this simpliο¬es down to y = c 2. x-intercepts: Found by making y = 0, this means solving 0 = ax2 + bx + c 3. Vertex: Let x = β 2a to ο¬nd x. Then plug this value into the equation to ο¬nd b y. After ο¬nding these points |
we can connect the dots with a smooth curve to ο¬nd our graph! Example 502. y = x2 + 4x + 3 Find the key points y = 3 y = c is the y intercept β 0 = x2 + 4x + 3 To ο¬nd x intercept we solve the equation β 0 = (x + 3)(x + 1) x + 3 = 0 and x + 1 = 0 1 1 Our x Factor Set each factor equal to zero Solve each equation intercepts 1 3 and (1) 2)2 + 4 To ο¬nd the vertex, ο¬rst use x = β 2a = β 2) + 3 Plug this answer into equation to ο¬nd y 8 + 3 Evaluate 1 The y 1) Vertex as a point coordinate β β y = 2, β ( β β β coordinate β 381 Graph the y-intercept at 3, the x3 and intercepts at 1, and the vertex at ( 1). Connect the dots with a smooth curve in a U shape to get our parabola. β β 2, β β Our Solution If the a in y = ax2 + bx + c is a negative value, the parabola will end up being an upside-down U. The process to graph it is identical, we just need to be very careful of how our signs operate. Remember, if a is negative, then ax2 will also be negative because we only square the x, not the a. Example 503. y = β 3x2 + 12x y = 9 9 β β Find key points intercept is y = c y β β 3)(x 3x2 + 12x 0 = β 3(x2 0 = β 3(x 0 = β 3 = 0 and x β 4x + 3) 1 and x = 1 Our x β β + 3 + 3 x 9 To ο¬nd x intercept solve this equation β Factor out GCF ο¬rst, then factor rest Set each factor with a varaible equal to zero Solve each equation intercepts β y = 12 3) = β β 12 x = β 6 2( β 3(2)2 + 12(2) 3(4) + 24 12 + 24 β y = β y = β b = 2 To ο¬nd the vertex, ο¬rst use x = β 2a 9 Plug this value into equation to ο¬nd y β 9 |
Evaluate β 9 β y = 3 (2, 3) Vertex as a point value of vertex β y coordinate β Graph the y-intercept at 9, the xintercepts at 3 and 1, and the vertex at (2, 3). Connect the dots with smooth curve in an upside-down U shape to get our parabola. β Our Solution 382 It is important to remember the graph of all quadratics is a parabola with the same U shape (they could be upside-down). If you plot your points and we cannot connect them in the correct U shape then one of your points must be wrong. Go back and check your work to be sure they are correct! Just as all quadratics (equation with y = x2) all have the same U-shape to them and all linear equations (equations such as y = x) have the same line shape when graphed, diο¬erent equations have diο¬erent shapes to them. Below are some common equations (some we have yet to cover!) with their graph shape drawn. Absolute Value y = x | | Cubic y = x3 Quadratic y = x2 Exponential y = ax Square Root y = xβ Logarithmic y = logax 383 9.11 Practice - Graphs of Quadratics Find the vertex and intercepts of the following quadratics. Use this information to graph the quadratic. 1) y = x2 β 3) y = 2x2 2x 8 β 12x + 10 β 2) y = x2 β 4) y = 2x2 2x 3 β 12x + 16 β 18 45 2x2 + 12x β 3x2 + 24x β x2 + 4x + 5 5) y = 7) y = β β 9) y = β 11) y = β 5 x2 + 6x β 2x2 + 16x 13) y = β β 15) y = 3x2 + 12x + 9 24 17) y = 5x2 19) y = β β 5x2 40x + 75 60x 175 β β 10 9 β β 6) y = β 2x2 + 12x 3x2 + 12x 8) y = β 10) y = β x2 + 4x β 2x2 + 16x 3 30 β 12) y = β 14) y = 2x2 + 4x 6 β 16 |
) y = 5x2 + 30x + 45 18) y = 5x2 + 20x + 15 20) y = β 5x2 + 20x 15 β 384 Chapter 10 : Functions 10.1 Function Notation..................................................................................386 10.2 Operations on Functions........................................................................393 10.3 Inverse Functions...................................................................................401 10.4 Exponential Functions...........................................................................406 10.5 Logarithmic Functions...........................................................................410 10.6 Application: Compound Interest............................................................414 10.7 Trigonometric Functions........................................................................420 10.8 Inverse Trigonometric Functions............................................................428 385 10.1 Functions - Function Notation Objective: Idenο¬ty functions and use correct notation to evaluate functions at numerical and variable values. There are many diο¬erent types of equations that we can work with in algebra. An equation gives the relationship between variables and numbers. Examples of several relationships are below: 3)2 (x β 9 (y + 2)2 4 β = 1 and y = x2 β 2x + 7 and y + xβ 7 = xy β There is a speical classiο¬cation of relationships known as functions. Functions have at most one output for any input. Generally x is the variable that we plug into an equation and evaluate to ο¬nd y. For this reason x is considered an input variable and y is considered an output variable. This means the deο¬nition of a function, in terms of equations in x and y could be said, for any x value there is at most one y value that corresponds with it. A great way to visualize this deο¬nition is to look at the graphs of a few relationships. Because x values are vertical lines we will draw a vertical line through the graph. If the vertical line crosses the graph more than once, that means we have too many possible y values. If the graph crosses the graph only once, then we say the relationship is a function. 386 Example 504. Which of the following graphs are graphs of functions? line Drawing a vertical through this graph will only cross the graph once, it is a function. Drawing a vertical line through this graph will cross the graph twice, once at top and once at bottom. This is not a function. line Drawing a vertical through this graph will cross the graph only once, it is a function. We can look at the above idea in an |
algebraic method by taking a relationship and solving it for y. If we have only one solution then it is a function. Example 505. Is 3x2 y = 5 a function? 3x2 β 3x2 + 5 β 1 1 β 3x2 β y = 1 β y = 3x2 β 5 β β β Solve the relation for y Subtract 3x2 from both sides Divide each term by 1 β Only one solution for y. It is a function Yes! Example 506. Is y2 x = 5 a function? β + x + x y2 = x + 5 y2 = y = Β± Β± x + 5β x + 5β p Solve the relation for y Add x to both sides Square root of both sdies Simplify Two solutions for y (one +, one ) β No! Not a function Once we know we have a function, often we will change the notation used to emphasis the fact that it is a function. Instead of writing y =, we will use function notation which can be written f (x) =. We read this notation βf of xβ. So for 387 the above example that was a function, instead of writing y = 3x2 5, we could have written f (x) = 3x2 5. It is important to point out that f (x) does not mean f times x, it is mearly a notation that names the function with the ο¬rst letter (function f ) and then in parenthesis we are given information about what variables are in the function (variable x). The ο¬rst letter can be anything we want it to be, often you will see g(x) (read g of x). β β World View Note: The concept of a function was ο¬rst introduced by Arab mathematician Sharaf al-Din al-Tusi in the late 12th century Once we know a relationship is a function, we may be interested in what values can be put into the equations. The values that are put into an equation (generally the x values) are called the domain. When ο¬nding the domain, often it is easier to consider what cannot happen in a given function, then exclude those values. Example 507. Find the domain: f (x) = 1 3x β x2 + x 6 β 6 0 β 2 x2 + x (x + 3)(x x + 3 |
0 and x 3 3 β β x β 3, 2 Our Solution With fractions, zero canβ²t be in denominator Solve by factoring Set each factor not equal to zero Solve each equation The notation in the previous example tells us that x can be any value except for 3 and 2. If x were one of those two values, the function would be undeο¬ned. β Example 508. Find the domain: f (x) = 3x2 x With this equation there are no bad values β All Real Numbers or R Our Solution In the above example there are no real numbers that make the function undeο¬ned. This means any number can be used for x. Example 509. Find the domain: f (x) = 2x β Square roots canβ²t be negative 3 β 388 2x 3 > 0 β + 3 + 3 2x > 3 2 2 x > 3 2 Set up an inequality Solve Our Solution The notation in the above example states that our variable can be 3 number larger than 3 ο¬ned (without using imaginary numbers). 2 or any 2. But any number smaller would make the function unde- Another use of function notation is to easily plug values into functions. If we want to substitute a variable for a value (or an expression) we simply replace the variable with what we want to plug in. This is shown in the following examples. Example 510. f (x) = 3x2 β 2) = 3( f ( β f ( β 4x; ο¬nd f ( 2)2 4( β 4( 2) = 3(4) f ( β β β β β 2 in for x in the function Substitute 2) 2) Evaluate, exponents ο¬rst 2) Multiply β 2) = 12 + 8 Add β f ( 2) = 20 Our Solution β Example 511. h(x) = 32x β 6; ο¬nd h(4) 6 h(4) = 32(4) h(4) = 38 β 6 β Substitute 4 in for x in the function Simplify exponent, mutiplying ο¬rst Subtract in exponent h(4) = 32 Evaluate exponent h(4) = 9 Our Solution Example 512. k(a) = 2 | k( a + 4 ; ο¬nd k( | 7) = 2 β k( | β 7) = 2 7( |
β 7) = 2(3) Multiply 7 in for a in the function Substitute β Add inside absolute values Evaluate absolute value 389 k( β 7) = 6 Our Solution As the above examples show, the function can take many diο¬erent forms, but the pattern to evaluate the function is always the same, replace the variable with what is in parenthesis and simplify. We can also substitute expressions into functions using the same process. Often the expressions use the same variable, it is important to remember each variable is replaced by whatever is in parenthesis. Example 513. g (x) = x4 + 1; ο¬nd g(3x) Replace x in the function with (3x) g(3x) = (3x)4 + 1 g(3x) = 81x4 + 1 Our Solution Simplify exponet Example 514. p(t) = t2 β t; ο¬nd p(t + 1) Replace each t in p(t) with (t + 1) Square binomial p(t + 1) = (t + 1)2 p(t + 1) = t2 + 2t + 1 β β p(t + 1) = t2 + 2t + 1 (t + 1) (t + 1) Distribute negative 1 Combine like terms t β β p(t + 1) = t2 + t Our Solution It is important to become comfortable with function notation and how to use it as we transition into more advanced algebra topics. 390 10.1 Practice - Function Notation Solve. 1) Which of the following is a function? a) c) e) y = 3x 7 β g) yβ + x = 2 b) d) f) y2 x2 = 1 β h) x2 + y2 = 1 Specify the domain of each of the following funcitons. 5x + 1 2) f (x) = β 4) s(t) = 1 t2 6) s(t) = 1 t2 + 1 2 8) f (x) = β x2 3x β 10 y(x) = x x2 β 25 4 β β 3) f (x) = 5 4x β 5) f (x) = x2 β 7) f (x) = x β 3x 4 β 16 β β 9) h(x) = 3x β x |
2 β 12 25 391 Evaluate each function. 11) g(x) = 4x 4; Find g(0) β 3x + 1 13) f (x) = | + 1; Find f (0) | 17) f (t) = 3t 19) f (t) = | β t + 3 2) 2; Find f ( β ; Find f (10) | 15) f (n; Find f ( 6) β 16) f (n) = n 12) g(n) = 3 5β n; Find g(2) β 14) f (x) = x2 + 4; Find f ( Β· 9) β 3; Find f (10) β 1 β 3a 18) f (a) β 3; Find f (2) β 20) w(x) = x2 + 4x; Find w( 5) β 4x + 3; Find w(6) 21) w(n) = 4n + 3; Find w(2) 22) w(x) = β 23) w(n) = 2n+2; Find w( 2) 25) p(n) = n 3 | β β ; Find p(7) | 27) p(t) = t3 + t; Find p(4) β n 29) k(n) = 1 ; Find k(3) | 31) h(x) = x3 + 2; Find h( β | 33) h(x) = 3x + 2; Find h( 4x) 1 + x) β β 35) h(t) = 2 3t β β 1 + 2; Find h(n2) 37) g(x) = x + 1; Find g(3x) 39) g(x) = 5x; Find g( 3 β β x) 24) p(x) = x + 1; Find p(5) β | 26) k(a) = a + 3; Find k( | 1) β 2; Find k(2) Β· 42t+1 + 1; Find p( 28) k(x) = 2 β 42x β 30) p(t) = 2 β Β· 2) β 32) h(n) = 4n + 2; Find h(n + 2) 2a+3; Find h( a 4 ) 34) h(a) = 3 β 36) |
h(x) = x2 + 1; Find h( x 4 ) Β· 38) h(t) = t2 + t; Find h(t2) 40) h(n) = 5n β 1 + 1; Find h( n 2 ) 392 10.2 Functions - Operations on Functions Objective: Combine functions using sum, diο¬erence, product, quotient and composition of functions. Several functions can work together in one larger function. There are 5 common operations that can be performed on functions. The four basic operations on functions are adding, subtracting, multiplying, and dividing. The notation for these functions is as follows. Addition Subtraction Multiplication Division (f + g)(x) = f (x) + g(x) (f g(x) g)(x) = f (x) β β g)(x) = f (x)g(x) (f Β· f f (x) g g(x) (x) = When we do one of these four basic operations we can simply evaluate the two functions at the value and then do the operation with both solutions Example 515. f ( β x f (x) = x2 β g(x) = x + 1 ο¬nd (f + g)( 2 β 3) β Evaluate f and g at 3 β 3) = ( 3)2 β f ( β ( 3) 3) = 9 + 3 β β 2 Evaluate f at β 2 β 3) = 10 f ( β g( β 3) = ( g( β 3) + 1 Evaluate g at β 3) + g( (10) + ( β β 3) Add the two functions together 2) Add 8 Our Solution The process is the same regardless of the operation being performed. Example 516. h(x) = 2x k(x) = Find (h 4 β 3x + 1 k)(5) β Β· Evaluate h and k at 5 h(5) = 2(5) β 4 Evaluate h at 5 393 h(5) = 10 4 h(5) = 6 β k(5) = 3(5) + 1 Evaluate k at 5 β k(5) = β k(5) = 15 + 1 14 β h(5)k(5) Multiply the two results together (6)( 14) Multiply 84 Our Solution β β Often as we add, subtract, |
multiply, or divide functions, we do so in a way that keeps the variable. If there is no number to plug into the equations we will simply use each equation, in parenthesis, and simplify the expression. Example 517. f (x) = 2x 4 β g(x) = x2 Find (f β β x + 5 Write subtraction problem of functions g)(x) (2x β 2x 4) β f (x) (x2 g(x) Replace f (x) with (2x β β x + 5) Distribute the negative 5 Combine like terms 9 Our Solution β 4 β x2 + x β x2 + 3x β β β 3) and g(x) with (x2 x + 5) β The parenthesis are very important when we are replacing f (x) and g(x) with a variable. In the previous example we needed the parenthesis to know to distribute the negative. Example 518. 5 β Write division problem of functions f (x) = x2 g(x) = x f g Find 4x β 5 β (x) f (x) g(x) Replace f (x) with (x2 4x β β 5) and g(x) with (x 5) β (x2 β (x 4x β β 5) 5) To simplify the fraction we must ο¬rst factor 394 (x 5)(x + 1) 5) β (x β Divide out common factor of x 5 β x + 1 Our Solution Just as we could substitute an expression into evaluating functions, we can substitute an expression into the operations on functions. Example 519. f (x) = 2x 1 β g(x) = x + 4 Write as a sum of functions Find (f + g)(x2) f (x2) + g(x2) Replace x in f (x) and g(x) with x2 1] + [(x2) + 4] Distribute the + does not change the problem [2(x2) β 2x2 β 1 + x2 + 4 Combine like terms 3x2 + 3 Our Solution Example 520. f (x) = 2x 1 β g(x) = x + 4 Write as a product of functions Find (f g)(3x) Β· [2(3x) (6x 18x2 + 24x β β f (3x)g( |
3x) Replace x in f (x) and g(x) with 3x 1][(3x) + 4] Multiply our 2(3x) 1)(3x + 4) FOIL 3x 18x2 + 21x 4 Combine like terms 4 Our Solution β β β The ο¬fth operation of functions is called composition of functions. A composition of functions is a function inside of a function. The notation used for composition of functions is: (f β¦ g)(x) = f (g(x)) To calculate a composition of function we will evaluate the inner function and substitute the answer into the outer function. This is shown in the following example. 395 Example 521. a(x) = x2 2x + 1 β b(x) = x β b)(3) Find (a β¦ 5 Rewrite as a function in function a(b(3)) Evaluate the inner function ο¬rst, b(3) 2 This solution is put into a, a( 5 = 2) β b(3) = (3) 2)2 β 2( β a( β 2) = ( 2) + 1 Evaluate β a( β β 2) = 4 + 4 + 1 Add β a( β 2) = 9 Our Solution 396 We can also evaluate a composition of functions at a variable. In these problems we will take the inside function and substitute into the outside function. Example 522. x f (x) = x2 β g(x) = x + 3 Find (f g)(x) β¦ Rewrite as a function in function f (g(x)) Replace g(x) with x + 3 f (x + 3) Replace the variables in f with (x + 3) (x + 3)2 (x2 + 6x + 9) β β x2 + 6x + 9 (x + 3) Evaluate exponent (x + 3) Distribute negative 3 Combine like terms x β β x2 + 5x + 6 Our Solution f )(x) as the It is important to note that very rarely is (f following example will show, using the same equations, but compositing them in the opposite direction. g)(x) the same as (g β¦ β¦ Example 523. f (x) = x2 x β g(x) = x + 3 Find (g f )(x) β¦ Rewrite as |
a function in function g(f (x)) Replace f (x) with x2 x β g(x2 x) Replace the variable in g with (x2 x) β x) + 3 Here the parenthesis donβ²t change the expression x + 3 Our Solution β (x2 x2 β β World View Note: The term βfunctionβ came from Gottfried Wihelm Leibniz, a German mathematician from the late 17th century. 397 10.2 Practice - Operations on Functions Perform the indicated operations. 1) g(a) = a3 + 5a2 f(a) = 2a + 4 Find g(3) + f (3) 3) g(a) = 3a + 3 f (a) = 2a 2 Find (g + f )(9) β 5) g(x) = x + 3 f(x) = Find (g x + 4 f )(3) β β 7) g(x) = x2 + 2 f (x) = 2x + 5 Find (g f )(0) β 9) g(t) = t 3 β 3t3 + 6t h(t) = β Find g(1) + h(1) 11) h(t) = t + 5 g(t) = 3t 5 β g)(5) Find (h Β· 13) h(n) = 2n g(n) = 3n Find h(0) β β Γ· 2a g(a) = a2 + 3 Find ( f g )(7) 15) f (a) = β 1 5 g(0) 4 β 17) g(x) = β h(x) = 4x Find (g x3 2 β h)(x) β 19) f (x) = 3x + 2 β g(x) = x2 + 5x Find (f g)(x) β 21) g(x) = 4x + 5 h(x) = x2 + 5x Find g(x) h(x) Β· 2) f (x) = 3x2 + 3x β g(x) = 2x + 5 Find f ( 4) g( 4) β Γ· β 4) g(x) = 4x + 3 h(x) = x3 Find (g β h)( 2x2 β β 4x + 1 6) g(x |
) = β h(x) = 2x 1 β Find g(5) + h(5) β 1) 8) g(x) = 3x + 1 f (x) = x3 + 3x2 f (2) Find g(2) Β· 10) f (n) = n 5 g(n) = 4n + 2 Find (f + g)( β 12) g(a) = 3a 2 h(a) = 4a 2 Find (g + h) ( β β 8) β 10) β 14) g(x) = x2 2 h(x) = 2x + 5 Find g( β β 16) g(n) = n2 h(n) = 2n Find (g β 18) g(x) = 2x h(x) = x3 Find (g β 6) + h( 6) β 3 β 3 β h)(n) 3 2x2 + 2x β β h)(x) 4 20) g(t) = t h(t) = 2t Find (g β h)(t) Β· 22) g(t) = 2t2 h(t) = t + 5 Find g(t) β 5t β h(t) Β· 398 23) f (x) = x2 5x β g(x) = x + 5 Find (f + g)(x) 25) g(n) = n2 + 5 f (n) = 3n + 5 Find g(n) 27) g(a) = β f (a) = 3a + 5 Find ( g f )(a) f (n) Γ· 2a + 5 29) h(n) = n3 + 4n g(n) = 4n + 5 Find h(n) + g(n) 31) g(n) = n2 4n β h(n) = n 5 β h(n2) Find g(n2) Β· 33) f (x) = 2x g(x) = β β Find (f + g)( 3x 1 β x) 4 β 35) f (t) = t2 + 4t g(t) = 4t + 2 Find f (t2) + g(t2) 37) g(a) = a3 + 2a h(a) = 3a + 4 Find ( g x) h )( 39) |
f (n) = β g(n) = 2n + 1 Find (f β 3n2 + 1 g)( n 3 ) β 4x + 1 41) f (x) = g(x) = 4x + 3 Find (f g)(9) β β¦ 43) h(a) = 3a + 3 g(a) = a + 1 Find (h g)(5) β¦ 45) g(x) = x + 4 h(x) = x2 1 β h)(10) Find (g β¦ 24) f (x) = 4x g(x) = 3x2 Find (f + g)(x) β β 4 5 26) f (x) = 2x + 4 g(x) = 4x 5 g(x) Find f (x) β β 28) g(t) = t3 + 3t2 h(t) = 3t Find g(t) 5 h(t) β β 30) f (x) = 4x + 2 g(x) = x2 + 2x Find f (x) Γ· 32) g(n) = n + 5 h(n) = 2n Find (g β h)( Β· g(x) 5 β 3n) 34) g(a) = 2a β h(a) = 3a Find g(4n) h(4n) Γ· 36) h(n) = 3n 2 β 3n2 g(n) = β Find h( n 3 ) 4n β g( n 3 ) Γ· 4x + 2 5 38) g(x) = β h(x) = x2 β Find g(x2) + h(x2) 40) f (n) = 3n + 4 g(n) = n3 5n β g( n Find f ( n 2 ) 2 ) β 42) g(x) = x Find (g 1 β g)(7) β¦ 44) g(t) = t + 3 h(t) = 2t 5 β h)(3) Find (g β¦ 46) f (a) = 2a 4 g(a) = a2 + 2a g)( Find (f β β¦ 4) β 399 47) f (n) = 4n + 2 β g(n) = n + 4 Find (f g)(9) β¦ 49) g |
(x) = 2x 4 β h(x) = 2x3 + 4x2 Find (g h)(3) β¦ 51) g(x) = x2 5x h(x) = 4x + 4 Find (g h)(x) β 53) f (a) = β g(a) = 4a Find (f 2a + 2 g)(a) β¦ β¦ 55) g(x) = 4x + 4 f (x) = x3 1 β f )(x) Find (g β¦ 57) g(x) = x + 5 3 β f )(x) β f (x) = 2x Find (g β¦ 59) f (t) = 4t + 3 g(t) = β Find (f β¦ 4t 2 β g)(t) 48) g(x) = 3x + 4 h(x) = x3 + 3x h)(3) Find (g β¦ 50) g(a) = a2 + 3 g)( Find (g β¦ 3) β 52) g(a) = 2a + 4 4a + 5 h)(a) h(a) = β Find (g β¦ t 4 β g)(t) β Find (g β¦ 54) g(t) = 4n β 56) f (n) = 2n2 g(n) = n + 2 Find (f β g)(n) β¦ 58) g(t) = t3 f (t) = 3t Find (g t β 4 β f )(t) β¦ 60) f (x) = 3x 4 β g(x) = x3 + 2x2 g)(x) Find (f β¦ 400 10.3 Functions - Inverse Functions Objective: Identify and ο¬nd inverse functions. When a value goes into a function it is called the input. The result that we get when we evaluate the function is called the output. When working with functions sometimes we will know the output and be interested in what input gave us the output. To ο¬nd this we use an inverse function. As the name suggests an inverse function undoes whatever the function did. If a function is named f (x), the 1(x) (read βf inverse of xβ). The negative one is inverse function will be named f β not an exponent, but mearly a symbol |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.