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Unformatted text preview: Certificate Mathematics in Action Full Solutions 4A 4 More about Equations Followup Exercise p. 162 1. By substituting x 2 = u into the equation x 4 10 x 2 + 9 = 0, we have u 2 10 u + 9 = 0 ( u 1)( u 9) = 0 u = 1 or u = 9 x 2 = u x 2 = 1 or x 2 = 9 x = 1 or x = 3 The real roots of the equation are 3, 1, 1 and 3. 2. By substituting x 2 = u into the equation x 4 + 3 x 2 4 = 0, we have u 2 + 3 u 4 = 0 ( u 1)( u + 4) = 0 u = 1 or u = 4 x 2 = u x 2 = 1 or x 2 = 4 (rejected) x = 1 The real roots of the equation are 1 and 1. 3. By substituting x 2 = u into the equation 4 x 4 17 x 2 + 4 = 0, we have 4 u 2 17 u + 4 = 0 (4 u 1)( u 4) = 0 u = 4 1 or u = 4 x 2 = u x 2 = 4 1 or x 2 = 4 x = 2 1 or x = 2 The real roots of the equation are 2, 2 1 , 2 1 and 2. 4. By substituting x 3 = u into the equation x 6 + 2 x 3 + 1 = 0, we have u 2 + 2 u + 1 = 0 ( u + 1) 2 = 0 u = 1 x 3 = u x 3 = 1 x = 1 The real root of the equation is 1. p. 163 1. 8 4 2 x x = 2 2 2 8 4  x x = 2 2 x 2 4 x 8 = 4 x 2 4 x 12 = 0 ( x + 2)( x 6) = 0 x = 2 or x = 6 Checking: When x = 2, 8 4 2 x x = 8 ) 2 ( 4 ) 2 ( 2  = 2 When x = 6, 8 4 2 x x = 8 6 4 6 2  = 2 The real roots of the equation are 2 and 6. 2. x + x 3 = 4 x 4 = x 3 ( x 4) 2 = ( 29 2 3 x x 2 8 x + 16 = 9 x x 2 17 x + 16 = 0 ( x 1)( x 16) = 0 x = 1 or x = 16 Checking: When x = 1, x + x 3 = 1 + 1 3 = 4 When x = 16, x + x 3 = 16 + 16 3 = 28 4 The real root of the equation is 1. Alternative Solution By substituting x = u into the equation x + x 3 = 4, we have u 2 + 3 u = 4 u 2 + 3 u 4 = 0 ( u 1)( u + 4) = 0 u = 1 or u = 4 x = u x = 1 or x = 4 (rejected) x = 1 The real root of the equation is 1. 3. x 2 3 x = 0 x = 2 3 x x 2 = ( 29 2 2 3 x x 2 = 9 x 18 x 2 9 x + 18 = 0 ( x 3)( x 6) = 0 x = 3 or x = 6 Checking: When x = 3, x 2 3 x = 3 2 3 3 = 0 When x = 6, x 2 3 x = 6 2 6 3 = 0 The real roots of the equation are 3 and 6. 4. x + 2 2 x = 10 2 2 x = 10 x ( 29 2 2 2 x = (10 x ) 2 4 x 8 = 100 20 x + x 2 x 2 24 x + 108 = 0 ( x 6)( x 18) = 0 x = 6 or x = 18 Checking: 83 4 More about Equations When x = 6, x + 2 2 x = 6 + 2 6 2 = 10 When x = 18, x + 2 2 x = 18 + 2 18 2 = 26 10 The real root of the equation is 6. p.168 For questions 1 to 4, refer to the graph below: 1. y = 3 x 5 x 1 2 3 y 2 1 4 The two graphs intersect at only one point (2, 1)....
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 Spring '09
 Math, Equations

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