1. Chapter 10 Section 3 Solving Quadratic Equations by the Quadratic Formula

2. Learning Objectives Solve quadratic equation be the quadratic formula Determine the number of solutions to a quadratic equation using the discriminant

3. Key Vocabulary Quadratic formula Discriminant

4. Quadratic Formula Standard form of a quadratic equation is ax 2 + bx + c = 0 where a is the coefficient of the squared term and b is the coefficient of the first-degree term and c is the constant. It is important to label the a, b, and c with the correct sign when substituting into the quadratic formula.

5. Solve the Quadratic Equation by Factoring x 2 + 7x + 10 = 0 (x + 2)(x + 5) = 0 x + 2 = 0 and x + 5 = 0 x = -2 and x = -5 Not all equations can be solve by factoring.

6. Solve the Quadratic Equation by Completing the Square x 2 + 7x + 10 = 0 ( ½ )(7) = 3.5 (3.5) 2 = 12.25 x 2 + 7x = -10 x 2 + 7x + 12.25 = -10 + 12.25 x 2 + 7x + 12.25 = 2.25 (x + 3.5) 2 = 2.25 x + 3.5 = ±1.5 x = -3.5 ±1.5 x = -3.5+1.5 and x = -3.5 – 1.5 x = -2 and x = -5 Perfect Square Trinomial

7. To Solve a Quadratic Equation by the Quadratic Formula 1.Write the equation in standard form to determine the values of a, b, and c. 2.Substitute the values for a, b and c from the equation in standard form into the quadratic formula and evaluate. 3.Check by placing back into original equation.

8. Solve using the Quadratic Formula x 2 + 7x + 10 = 0 Already in standard form a = 1 b = 7 c = 10 x = -2 x = -5

9. Solve using the Quadratic Formula 18x 2 – 3x – 1 = 0 Already in standard form a = 18 b = -3 c = -1

10. Solve using the Quadratic Formula 5n 2 + 2n - 1 = 0 Already in standard form a = 5 b = 2 c = -1

11. x 2 = 4x – 1 Put in standard form x 2 – 4x + 1 = 0 a = 1 b = -4 c = 1 Solve using the Quadratic Formula

12. 4x 2 = 2x – 3 Put in standard form 4x 2 – 2x + 3 = 0 a = 4 b = -2 c = 3 Solve using the Quadratic Formula < 0

13. m 2 = 64 Put in standard form m 2 +0x – 64 = 0 a = 1 b = 0 c = -64 Solve using the Quadratic Formula

14. x 2 +14x +45 = 0 Already in standard form a = 1 b = 14 c = 45 x = -5 x = -9 Solve using the Quadratic Formula

15. 12x 2 - 4x - 1 = 0 Already in standard form a = 12 b = -4 c = -1 Solve using the Quadratic Formula

16. 3x 2 + 4x - 8 = 0 Already in standard form a = 3 b = 4 c = -8 Solve using the Quadratic Formula

17. x 2 = 8x – 6 Put in standard form x 2 – 8x + 6 = 0 a = 1 b = -8 c = 6 Solve using the Quadratic Formula

18. a 2 – 121 = 0 Put in standard form a 2 + 0x – 121 = 0 a = 1 b = 0 c = -121 Solve using the Quadratic Formula

19. 4x 2 - 2x + 3 = 0 Already in standard form a = 4 b = -2 c = 3 No real solution Solve using the Quadratic Formula < 0

20. The length of a rectangle is 1 ft more than three times the width. Find the dimensions of the rectangle if the area is 30 ft 2, find the length and width A = LW Let x = width length = 3x + 1 a = 3 (3x + 1)x = 30b = 1 c = -30 3x 2 + x = 30 put in standard form 3x 2 + x – 30 = 0 W = 3 L = (3)(3) + 1 = 10 Solve an Application Problem using the Quadratic Formula

21. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant The expression under the radical is called the discriminant and can be used to determine the number of solutions b 2 – 4ac is called the discriminant b 2 – 4ac > 0 two distinct real number solutions b 2 – 4ac = 0one real number solution b 2 – 4ac < 0no real number solution, negative

22. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x 2 + 3x = 5b 2 – 4ac (3) 2 – (4)(2)(-5) Put in standard form9 – (8)(-5) 9 + 40 2x 2 + 3x – 5 = 049 a = 2 49 > 0 b = 3 c = -5Two distinct real number solutions

23. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x 2 + x + 3 = 0b 2 – 4ac (1) 2 – (4)(2)(3) Already in standard form1 – (8)(3) 1 – 24 a = 2 -23 b = 1 c = 3-23 < 0 No real number solutions

24. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 9x 2 - 24 x + 16 = 0b 2 – 4ac (-24) 2 – (4)(9)(16) Already in standard form256 – (36)(16) 576 – 576 a = 9 0 b = -24 c = 160 = 0 One real number solutions

25. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x 2 - 3 x - 7 = 0b 2 – 4ac (-3) 2 – (4)(2)(-7) Already in standard form9 – (8)(-7) 9 + 56 a = 2 65 b = -3 c = -765 > 0 Two distinct real number solutions

26. Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 3x 2 = - x + 4b 2 – 4ac (1) 2 – (4)(3)(-4) Put in standard form1 – (12)(-4) 1 – 48 3x 2 + x – 4 = 0-47 a = 3 -47 < 0 b = 1 c = -4No real number solutions

27. Remember The equation should be in standard form so that you can correctly determine a, b, and c. Make sure that you label a, b, and c with the correct sign. Check your answers by placing them back into the original equation. Use the discriminant to determine the number of solutions

28. HOMEWORK 10.3 Page 609 – 610: # 27, 29, 35, 36, 43, 47, 51, 59