1 / 37

3.5 Quadratic Equations

3.5 Quadratic Equations. OBJ: To solve a quadratic equation by factoring. DEF:  Standard form of a quadratic equation. ax 2 + bx + c = 0 NOTE:  Each equation contains a polynomial of the second degree. DEF:  Zero – product property. If mn = 0, then m = 0 or n = 0 or both = 0

hoyt-terry
Download Presentation

3.5 Quadratic Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 3.5 Quadratic Equations OBJ: To solve a quadratic equation by factoring

  2. DEF: Standard form of a quadratic equation ax2 + bx + c = 0 • NOTE:  Each equation contains a polynomial of the second degree.

  3. DEF: Zero – product property If mn = 0, then m = 0 or n = 0 or both = 0 • NOTE:  Solve some quadratic equations by: • Writing equation in standard form • Factoring • Setting each factor equal to 0

  4. P 68 EX 1: 3 c2 – 10c – 8 = 0 3 2 3 4 1 4 1 2 +2 3 2 • 4 -12 (3c + 2)(c – 4) = 0 c = - 2/3, 4

  5. P 68 EX 2: 5x = 6 – x 2 x 2 + 5x – 6 = 0 (x + 6)(x – 1) = 0 x = - 6, 1

  6. P 69 EX 3: – 7 x 2 = 21x 7 x 2 + 21x = 0 7x (x + 3) = 0 x = 0, - 3

  7. P 69 EX 3: 25 = 9 n 2 9 n 2 – 25 = 0 (3n – 5)(3n + 5) = 0 n = ± 5/3

  8. EX 5:5 c 2 + 7c – 6 = 0 • 3 5 2 • 2 1 3 -3 • 3 • 2 +10 (5c – 3)(c + 2) = 0 c = 3/5, -2

  9. EX 6: 7t = 20 – 3 t 2 3 t 2 + 7t – 20 = 0 • 4 3 5 • 5 1 4 -5 • 5 • 4 +12 (3t – 5)(t + 4) = 0 t = 5/3, -4

  10. EX 7:36 = 25 x 2 25 x 2 – 36 = 0 (5x – 6)(5x + 6) = 0 x = ± 6/5

  11. EX 8:–2 x 2 = 5x 2 x 2 + 5x = 0 2x(x + 5) = 0 x = 0, - 5

  12. P69 EX 5:3 n 2 – 15n + 18 = 0 3 (n 2 – 5n + 6) = 0 3 (n – 3)(n – 2) = 0 n = 3, 2

  13. EX 10:7 n 2 + 14n – 56 = 0 7 (n 2 + 2n – 8) = 0 7 (n + 4)(n – 2) = 0 n = - 4, 2

  14. P69 EX 4:x 4 – 13 x 2 + 36 = 0 (x 2 – 9)(x 2 – 4) = 0 (x – 3)(x + 3)(x – 2)(x + 2) =0 x = ± 3, ± 2

  15. EX 12:y 4 – 5 y 2 + 4 = 0 (y 2 – 4)(y 2 – 1) = 0 (y – 2)(y + 2)(y – 1)(y + 1) = 0 Y = ± 2, ± 1

  16. EX 13:y 4 – 10 y 2 + 9 = 0 (y 2 – 9)(y 2 – 1) = 0 (y – 3)(y + 3)(y – 1)(y + 1) = 0 Y = ± 3, ± 1

  17. EX 14:y 4 = 20 – y 2 y 4 + y 2 – 20 = 0 (y 2 + 5)(y 2 – 4) = 0 (y 2 + 5)(y – 2)(y + 2) = 0 Y = ± i√ 5, ± 2

  18. EX 15:y 4 = 12 + y 2 y 4 – y 2 – 12 = 0 (y 2 – 4)(y 2 + 3) = 0 (y – 2)(y + 2)(y 2 + 3) = 0 Y = ± 2, ± i√ 3

  19. 6.1 Square Roots OBJ:To solve a quadratic equation by using the definition of square root DEF:Square root If x 2 = k, then x = ±√k, for k ≥ 0

  20. P 139 EX 1: x 2 + 5 = 15 x 2 = 10 x = ± √10

  21. P 139 EX 1:3 y 2 = 75 y 2 = 25 y = ± 5

  22. EX 3: 6 y 2 – 20 = 8 – y 2 7y 2 = 28 y 2 = 4 y = ± 2

  23. EX 4:3 n 2 + 9 = 7 n 2 – 35 44 = 4n 2 11 = n 2 ±√11 = n

  24. 7.3 The Quadratic Formula OBJ:To solve a quadratic equation by using the quadratic formula DEF: The quadratic formula x = -b ± √b2 – 4ac 2a

  25. P169 EX 1:3 x 2 + 5x – 4 = 0 x = -5 ± √52 – 4(3)(-4) 2(3) = -5 ± √25 + 48 6 x = -5 ± √73 6

  26. 4 x 2 – 4 x – 11 = 0 x = -(-4)±√(-4)2 – 4(4)(-11) 2(4) x = 4 ± √16 + 176 8 = 4 ± √192 8 = 4 ± 8√3 8 = 4(1 ± 2√3 8 = 4(1 ± 2√3 8 2 = 1 ± 2√3 2 P170 EX 2 :4 x 2 = 11 + 4x

  27. P 170 EX 3:5 x 2 – 9x = 0 x(5x – 9) = 0 x = 0, 9/5

  28. P 170 EX 3:y 2 – 150 = 0 y 2 = 150 y = ± √150 = ± 5√6

  29. EX 5: 4 x 2 – 7x + 2 = 0 x = -(-7) ± √(-7)2 – 4(4)(2) 2(4) = 7 ± √49 – 32 8 = 7 ± √17 8

  30. 9 x 2 – 12x + 1 = 0 x = -(-12)±√(-12)2 – 4(9)(1) 2(9) = 12 ± √144– 36 18 = 12 ± √108 18 = 12 ± 6√3 18 = 12 ± 6√3 18 = 6(2 ± √3) 18 3 = 2 ± √3 3 EX 6:9 x 2 = 12x – 1

  31. EX 7:6 x 2 + 5x = 0 x(6x + 5) = 0 x = 0, -5/6

  32. EX 8:72 – x 2 = 0 x 2 = 72 x = ± 6√2

  33. 8.3 Equations With Imaginary Number Solutions OBJ:To solve an equation whose solutions are imaginary

  34. 3 x 2 – 4x + 2 = 0 x = -(-4)±√(-4)2 – 4(3)(2) 2(3) = 4 ± √16– 24 6 = 4 ± √-8 6 = 4 ± 2i√2 6 = 2(2 ± i√2 6 3 = 2 ± i√2 3 P 193 EX 1:3 x 2 + 2 = 4x

  35. P193 EX 2:2 x 4 + 3 x 2 – 20 = 0 (2 x 4 – 5 )(x 2 + 4) = 0 x 2 = 5/2 or -4 x = ±√10/2 or ± 2i

  36. 2 x 2 – 6x + 7 = 0 x = -(-6)±√(-6)2 – 4(2)(7) 2(2) = 6 ± √36– 56 4 = 6 ± √-20 4 x = 6 ± 2i√5 4 = 2(3 ± i√5) 4 = 2(3 ± i√5) 4 2 = 3 ± i√5 2 EX 3:2 x 2 + 7 = 6x

  37. EX 4:27 – 6 y 2 = y 4 y 4 + 6 y 2 – 27 = 0 (y 2 + 9)(y 2 – 3) = 0 y = ± 3i, ± √3

More Related