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1 Procedure for Solving radical equaions 1. Algebraically isolae one radical by iself on one side of equal sign. 2. Raise each side of he equaion o an appropriae power o remove he radical. 3. Simplify. 4. If he equaion sill conains a radical, repea seps 1 hrough Once all he radicals are removed, solve he equaion. 6. Check all soluions, and exclude any ha do no saisfy he original equaion. These excluded soluions are called exraneous soluions. Example Solve 2 4x x x x + 9 (isolae he radical 4x + 1) ( ) ( 2 x + 4 4x + 1 (now square each side o remove radical) 2 (x + 4)2 4x + 1 (simplify) 4 16x + 4 (x + 4 (quadraic se i up for facoring) 16x + 4 x 2 + 8x + 16 x 2 8x (x 6)(x 2) 0 (x 6) 0 or (x 2) 0 x 6 or x 2 We are no done unil we have checked ha hese really are soluions: 2 4(6) (6) + 9 (check x 6) (True! x 6 is a soluion) 2 4(2) (2) + 9 (check x 2) (True! x 2 is a soluion) Insrucor: Barry McQuarrie Page 1 of 5

2 Example Solve 3x x x. 3x x x (we ve go a radical isolaed, so square boh sides) ( 3x x + 5 ( 7 2x (simplify) 3x x x + 4 x x (3x + 4)(x + 5) 1 3x (isolae he remaining radical) ( (3x + 4)(x + 5) ( 1 3x (square boh sides o remove he radical) (3x + 4)(x + 5) 1 + 6x + 9x 2 (simplify) 3x x x + 9x 2 (we have a quadraic, so se i up for facoring) 6x 2 13x 19 0 (facor by grouping: wo numbers whose produc is 114 sum is 13: 19, 6) 6x 2 19x + 6x 19 0 x(6x 19) + 1(6x 19) 0 (x + 1)(6x 19) 0 (x + 1) 0 or (6x 19) 0 x 1 or x 19 6 (bu we aren done ye!) We aren done unil we see if hese solve he original equaion: 3( 1) ( 1) ( 1) (check x 1) (True! So x 1 is a soluion) 3(19/6) (19/6) (19/6) (check x 19/6) 27/2 + 49/6 2/3 (False! So x 19/6 is no a soluion) The only soluion o he original radical equaion is x 1. Complex numbers a + bi where a is he real par and b is he imaginary par. i 2 1 or i 1 and i is called an imaginary number. For all a 0, a i a. (a + bi) + (c + di) (a + c) + (b + d)i (real par goes wih real par, imaginary par goes wih imaginary par). The complex conjugae of a + bi is equal o a bi. To divide wo complee numbers, muliply he numeraor and denominaor by he complex conjugae of he denominaor. The phrase imaginary number is a bi misleading he number exiss, i is jus somehing oher han a real number! Complex numbers are incredibly useful in a variey of areas of mahemaics. For now, you should see ha he real pars are colleced ogeher and he imaginary pars are colleced ogeher when you are working wih complex numbers. The basic fac ha i 2 1 allows you o simplify complex numbers. Insrucor: Barry McQuarrie Page 2 of 5

3 Example Divide he wo complex numbers: Use complex conjugae of he denominaor: i 6 3i. Variaion i 6 3i i 6 3i 6 + 3i 6 + 3i (7 + 14i)(6 + 3i) (6 3i)(6 + 3i) i + 21i + 42i2 36 9i i + 42( 1) (use i 2 1) 36 9( 1) 105i i (simplify) Direc variaion beween x and y means y kx, where k is he consan of variaion. Inverse variaion beween x and y means y k, where k is he consan of variaion. x In eiher case, use he given informaion o deermine he value of k. Then, use he equaion you have creaed o deermine he unknown quaniy. Join variaion jus means a quaniy depends on he produc of more han one quaniy. Example Police officers can use variaion o deec speeding. The speed of a car varies inversely wih he ime i akes o cover a cerain fixed disance. Beween wo poins on a highway, a car ravels 45 mph in 6 seconds. Wha is he speed of a car ha ravels he same disance in 9 seconds We are old here is an inverse variaion beween speed and ime: v k. where is ime in seconds and v is velociy in mph. Use he given informaion (car raveling a 45 mph covers he disance in 6 seconds) o deermine he value of he consan of variaion k: v k 45 k 6 k 270. Now we can answer he quesion asked abou he speed of a car ha covers he disance in 9 seconds: v 270 v v 30mph. Anoher ineresing quesion o ask is: If he speed limi is 65 mph, wha ime will a speeding car cover he disance v seconds. Any car ha covers he disance faser han 4.15 seconds is speeding. Insrucor: Barry McQuarrie Page 3 of 5

4 Solving Absolue Value Equaliies and Inequaliies: Three Cases Case1 For equaliies of he form ax + b cx + d, he soluion is ax + b cx + d or ax + b (cx + d). The soluion will be wo disinc numbers. Case2 For inequaliies of he form ax + b < c, where c > 0 he soluion is c < ax + b < c. NOTE: i is imporan ha he c is on he lef and he c is on he righ. If his isn he case, you will ge he wrong soluion. The soluion will be a se of poins beween wo numbers. Case3 For inequaliies of he form ax + b > c, where c > 0 he soluion is ax + b < c or ax + b > c. The soluion will be a se of numbers less han one number or greaer han anoher number. Example Solve he inequaliy 3x and skech he soluion on a number line. 3x x or 3x x 22 or 3x 26 x 22 3 or x 26 3 Example Solve he inequaliy 3x + 2 < 24 and skech he soluion on a number line. 3x + 2 < < 3x + 2 < < 3x < < x < 22 3 Insrucor: Barry McQuarrie Page 4 of 5

5 Example Solve he inequaliy 3x + 2 > 24 and skech he soluion on a number line. 3x + 2 > 24 3x + 2 > 24 or 3x + 2 < 24 3x > 22 or 3x < 26 x > 22 3 or x < 26 3 Insrucor: Barry McQuarrie Page 5 of 5

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