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  1. The Quadratic Formula 2-6 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. Warm Up Write each function in standard form. Evaluate b2 – 4ac for the given values of the valuables. 2.g(x) = 2(x + 6)2 – 11 1.f(x) = (x – 4)2 + 3 g(x) = 2x2 + 24x + 61 f(x) = x2 – 8x + 19 4.a = 1, b = 3, c = –3 3.a = 2, b = 7, c = 5 21 9

  3. Objectives Solve quadratic equations using the Quadratic Formula. Classify roots using the discriminant.

  4. Vocabulary discriminant

  5. You have learned several methods for solving quadratic equations: graphing, making tables, factoring, using square roots, and completing the square. Another method is to use the Quadratic Formula, which allows you to solve a quadratic equation in standard form. By completing the square on the standard form of a quadratic equation, you can determine the Quadratic Formula.

  6. Remember! To subtract fractions, you need a common denominator.

  7. The symmetry of a quadratic function is evident in the last step, . These two zeros are the same distance, , away from the axis of symmetry, ,with one zero on either side of the vertex.

  8. You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.

  9. Example 1: Quadratic Functions with Real Zeros Find the zeros of f(x)= 2x2 – 16x + 27 using the Quadratic Formula. 2x2– 16x + 27 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 2 for a, –16 for b, and 27 for c. Simplify. Write in simplest form.

  10. Example 1 Continued CheckSolve by completing the square. 

  11. Check It Out! Example 1a Find the zeros of f(x) = x2 + 3x –7 using the Quadratic Formula. x2 + 3x–7 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, 3 for b, and –7 for c. Simplify. Write in simplest form.

  12. Check It Out! Example 1a Continued CheckSolve by completing the square. x2 + 3x – 7 = 0 x2 + 3x = 7 

  13. Check It Out! Example 1b Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula. x2 – 8x + 10 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, –8 for b, and 10 for c. Simplify. Write in simplest form.

  14. Check It Out! Example 1b Continued CheckSolve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 16 = –10 + 16 (x + 4)2 = 6 

  15. Example 2: Quadratic Functions with Complex Zeros Find the zeros of f(x) = 4x2 + 3x + 2 using the Quadratic Formula. f(x)= 4x2 + 3x + 2 Set f(x) = 0. Write the Quadratic Formula. Substitute 4 for a, 3 for b, and 2 for c. Simplify. Write in terms of i.

  16. Check It Out! Example 2 Find the zeros of g(x) = 3x2 – x + 8 using the Quadratic Formula. Set f(x) = 0 Write the Quadratic Formula. Substitute 3 for a, –1 for b, and 8 for c. Simplify. Write in terms of i.

  17. The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

  18. Caution! Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.

  19. Example 3A: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 36 = 12x x2–12x + 36 = 0 b2 – 4ac (–12)2 – 4(1)(36) 144 – 144 = 0 b2 – 4ac = 0 The equation has one distinct real solution.

  20. Example 3B: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 40 = 12x x2–12x + 40 = 0 b2 – 4ac (–12)2 – 4(1)(40) 144 – 160 = –16 b2 –4ac < 0 The equation has two distinct nonreal complex solutions.

  21. Example 3C: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 30 = 12x x2–12x + 30 = 0 b2 – 4ac (–12)2 – 4(1)(30) 144 – 120 = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

  22. Check It Out! Example 3a Find the type and number of solutions for the equation. x2 – 4x = –4 x2–4x + 4 = 0 b2 – 4ac (–4)2 – 4(1)(4) 16 – 16 = 0 b2 – 4ac = 0 The equation has one distinct real solution.

  23. Check It Out! Example 3b Find the type and number of solutions for the equation. x2 – 4x = –8 x2–4x + 8 = 0 b2 – 4ac (–4)2 – 4(1)(8) 16 – 32 = –16 b2 – 4ac < 0 The equation has two distinct nonreal complex solutions.

  24. Check It Out! Example 3c Find the type and number of solutions for each equation. x2 – 4x = 2 x2–4x– 2 = 0 b2 – 4ac (–4)2 – 4(1)(–2) 16 + 8 = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

  25. The graph shows related functions. Notice that the number of real solutions for the equation can be changed by changing the value of the constant c.

  26. Example 4: Sports Application An athlete on a track team throws a shot put. The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t2 + 24.6t+ 6.5. The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t. To the nearest foot, how far does the shot put land from the athlete?

  27. Example 4 Continued Step 1 Use the first equation to determinehow long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t. y = –16t2 + 24.6t + 6.5 0 = –16t2 + 24.6t + 6.5 Set y equal to 0. Use the Quadratic Formula. Substitute –16 for a, 24.6 for b, and 6.5 for c.

  28. Example 4 Continued Simplify. The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.

  29. Example 4 Continued Step 2 Find the horizontal distance that the shot put will have traveled in this time. x = 29.3t x ≈ 29.3(1.77) Substitute 1.77 for t. x ≈ 51.86 Simplify. x ≈ 52 The shot put will have traveled a horizontal distance of about 52 feet.

  30. Example 4 Continued Check Use substitution to check that the shot put hits the ground after about 1.77 seconds. y = –16t2 + 24.6t + 6.5 y ≈ –16(1.77)2 + 24.6(1.77) + 6.5 y ≈ –50.13 + 43.54 + 6.5 y ≈ –0.09  The height is approximately equal to 0 when t = 1.77.

  31. Check It Out! Example 4 A pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of the water t seconds after its release is modeled by y = –16t2 – 2t + 500. the horizontal distance x in feet between the water and its point of release is modeled by x = 91t. The pilot’s altitude decreases, which changes the function describing the water’s height toy = –16t2 –2t + 400. To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?

  32. Check It Out! Example 4 Continued Step 1 Use the equation to determinehow long it will take the water to hit the ground. Set the height of the water equal to 0 feet, and then use the quadratic formula for t. y = –16t2 – 2t + 400 0 = –16t2 – 2t + 400 Set y equal to 0. Write the Quadratic Formula. Substitute –16 for a, –2 for b, and 400 for c.

  33. Check It Out! Example 4 Simplify. t ≈ –5.063 or t ≈ 4.937 The time cannot be negative, so the water lands on a target about 4.937 seconds after it is released.

  34. Check It Out! Example 4 Continued Step 2 The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. Find the horizontal distance that the water will have traveled in this time. x = 91t Substitute 4.937 for t. x ≈ 91(4.937) x ≈ 449.267 Simplify. x ≈ 449 The water will have traveled a horizontal distance of about 449 feet. Therefore, the pilot should start releasing the water when the horizontal distance between the helicopter and the fire is 449 feet.

  35. Check It Out! Example 4 Continued Check Use substitution to check that the water hits the ground after about 4.937 seconds. y = –16t2 – 2t + 400 y ≈ –16(4.937)2 – 2(4.937) + 400 y ≈ –389.983 – 9.874 + 400 y ≈ 0.143  The height is approximately equal to 0 when t = 4.937.

  36. Properties of Solving Quadratic Equations

  37. Properties of Solving Quadratic Equations

  38. Helpful Hint No matter which method you use to solve a quadratic equation, you should get the same answer.

  39. Lesson Quiz: Part I Find the zeros of each function by using the Quadratic Formula. 1. f(x) = 3x2 – 6x – 5 2. g(x) = 2x2 – 6x + 5 Find the type and member of solutions for each equation. 3. x2 – 14x + 50 4. x2 – 14x + 48 2 distinct real 2 distinct nonreal complex

  40. Lesson Quiz: Part II 5. A pebble is tossed from the top of a cliff. The pebble’s height is given by y(t) = –16t2+ 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t. How far will the pebble be from the base of the cliff when it hits the ground? about 18 ft