1. Section 1.2Quadratic Equations Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
2. Quadratic Equations Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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4. Solve the equation: x − 4x = 0 2 x( x − 4) = 0 x = 0 or x + 4 = 0 x = 0 or x − 4 The solution set is { 0, 4} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
5. Solve the equation: x2 = 6 − x x + x−6 = 0 2 ( x + 3)( x − 2) = 0 x = −3 or x = 2 The solution set is { −3, 2} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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7. Solve each equation. ( x + 3) 2 (a) x = 7 2 (b) =9 x+3= ± 9 x=± 7 x + 3 = ±3 x = 7 or x = − 7 x + 3 = 3 or x + 3 = −3 x = 0 or x = −6 {The solution set is − 7, 7 } The solution set is { 0, −6} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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13. Use the quadratic formula to find the real solutions if any, of the equation 2x − 4x +1 = 0 2 ax + bx + c = 0 2 b 2 − 4ac = (−4) 2 − 4(2)(1) = 16 − 8 = 8 b − 4ac > 0 so there are two real solutions 2 which can be found using the quadratic formula. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
14. Use the quadratic formula to find the real solutions if any, of the equation 2x − 4x +1 = 0 2 −b ± b − 4ac 2 x= 2a −(−4) ± 8 4± 8 2± 2 x= = = 2(2) 4 2 2 + 2 2 − 2    The solution set is  ,   2  2  Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
15. Use the quadratic formula to find the real solutions if any, of the equation 1 2 x − 6 x + 18 = 0 x 2 − 12 x + 36 = 0 2 ax 2 + bx + c = 0 b 2 − 4ac = (−12) 2 − 4(1)(36) = 144 − 144 = 0 b − 4ac = 0 so there is a repeated solution 2 which can be found using the quadratic formula. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
16. Use the quadratic formula to find the real solutions if any, of the equation 1 2 x − 6 x + 18 = 0 x − 12 x + 36 = 0 2 2 ax 2 + bx + c = 0 −b ± b 2 − 4acx= 2a −(−12) ± 0 12 x= = =6 The solution set is { 6} 2(1) 2 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
17. Use the quadratic formula to find the real solutions if any, of the equation 2x + 3 = 2x 2 Since b − 4ac < 0, 2 2x − 2x + 3 = 0 2 there is no real solution. ax 2 + bx + c = 0 b − 4ac = ( −2 ) − 4(2)(3) 2 2 = 4 − 24 = −20 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
18. Use the quadratic formula to find the real solutions if any, of the equation 1 2 −1 ± 49 −1 ± 7 6+ − 2 = 0 x= = x x 2(6) 12 6 x2 + x − 2 = 0 −1 + 7 1 = = 12 2 b − 4ac = ( 1) − 4(6)(−2) 2 2 −1 − 7 2 = =− = 1 + 48 = 49 12 3Since b 2 − 4ac > 0, 1 2 The solution set is  , − there are two real solutions. 2 3 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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20. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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22. 9 ( x − 18 ) = 144 ( x − 18) x − 18 = ±4 2 2 = 16 22 centimeters by 22 centimetersx = 18 ± 4 = 22 or 14 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
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