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Warm Up Write each function in standard form.

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

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Warm Up Write each function in standard form.

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  1. 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

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

  3. Vocabulary discriminant

  4. 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.

  5. 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.

  6. 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.

  7. 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.

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

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

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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?

  15. 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.

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

  17. 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.

  18. 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?

  19. 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.

  20. 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.

  21. 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.

  22. Properties of Solving Quadratic Equations

  23. Properties of Solving Quadratic Equations

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

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