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Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
Section 1.2 Quadratic Equations
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Section 1.2 Quadratic Equations

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Solving Quadratic Equations

Solving Quadratic Equations

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  • 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.
  • 3. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 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.
  • 6. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 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.
  • 8. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 9. Solve by completing the square: 2x2 + 6x − 5 = 0 2x + 6x = 5 2 3 19 x+ =± 2 4 2x2 6x 5 + = 2 2 2 19 3 x=± − 9 5 9 2 2x + 3x + = + 2 4 2 4 19 − 3 19 − 3 x= or − 2 2 2 3  19x+  = 2 4 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 10. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 11. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 12. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 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.
  • 19. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 20. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 21. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 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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