1 / 19

5.6 The Quadratic Formula and the Discriminant

5.6 The Quadratic Formula and the Discriminant. Algebra 2. Learning Target. Solve equations using the quadratic formula. Use the discriminant to determine the nature of the roots of a quadratic equation. Derive the quadratic formula from ax 2 + bx + c = 0 a ≠ 0.

dgalvan
Download Presentation

5.6 The Quadratic Formula and the Discriminant

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. 5.6 The Quadratic Formula and the Discriminant Algebra 2

  2. Learning Target • Solve equations using the quadratic formula. • Use the discriminant to determine the nature of the roots of a quadratic equation.

  3. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0 General form of a quadratic equation. Divide all by a Simplify Subtract c/a on both sides. Multiply by ½ and square the result.

  4. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0 Add the result to both sides. Simplify Multiply by common denominator Simplify

  5. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0 Square root both sides Simplify Common denominator/subtract from both sides Simplify

  6. Quadratic Formula • The solutions of a quadratic equation of the form ax2 + bx + c with a ≠ 0 are given by this formula: MEMORIZE!!!!

  7. Ex. 1: Solve t2 – 3t – 28 = 0 a = 1 b = -3 c = -28 There are 2 distinct roots—Real and rational.

  8. CHECK: t2 – 3t – 28 = 0 72 – 3(7) – 28 = 0 49 – 21 – 28 = 0 49 – 49 = 0  CHECK: t2 – 3t – 28 = 0 (-4)2 – 3(-4) – 28 = 0 16 + 12 – 28 = 0 28 – 28 = 0  Ex. 1: Solve t2 – 3t – 28 = 0

  9. Ex. 1: Solve t2 – 3t – 28 = 0 -- GRAPH

  10. Ex. 2: Solve x2 – 8x + 16 = 0 a = 1 b = -8 c = 16 There is 1 distinct root—Real and rational.

  11. Ex. 2: Solve x2 – 8x + 16 = 0 CHECK: x2 – 8x + 16 = 0 (4)2 – 8(4) + 16 = 0 16 – 32 + 16 = 0 32 – 32 = 0  There is 1 distinct root—Real and rational.

  12. Ex. 2: Solve Solve x2 – 8x + 16 = 0 -- GRAPH

  13. Ex. 3: Solve 3p2 – 5p + 9 = 0 a = 3 b = -5 c = 9 There is 2 imaginary roots.

  14. Ex. 3: Solve 3p2 – 5p + 9 = 0 NOTICE THAT THE PARABOLA DOES NOT TOUCH THE X-AXIS.

  15. Note: • These three examples demonstrate a pattern that is useful in determining the nature of the root of a quadratic equation. In the quadratic formula, the expression under the radical sign, b2 – 4ac is called the discriminant. The discriminant tells the nature of the roots of a quadratic equation.

  16. DISCRIMINANT • The discriminant will tell you about the nature of the roots of a quadratic equation.

  17. Ex. 4: Find the value of the discriminant of each equation and then describe the nature of its roots. 2x2 + x – 3 = 0 a = 2 b = 1 c = -3 b2 – 4ac = (1)2 – 4(2)(-3) = 1 + 24 = 25 The value of the discriminant is positive and a perfect square, so 2x2 + x – 3 = 0 has two real roots and they are rational.

  18. Ex. 5: Find the value of the discriminant of each equation and then describe the nature of its roots. x2 + 8 = 0 a = 1 b = 0 c = 8 b2 – 4ac = (0)2 – 4(1)(8) = 0 – 32 = – 32 The value of the discriminant is negative, so x2 + 8 = 0 has two imaginary roots.

  19. Pair-share • pp. 295 #16-45 even

More Related