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**The Quadratic Formula 5-6 Warm Up Lesson Presentation Lesson Quiz**

Holt McDougal Algebra 2 Holt Algebra 2

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**Write each function in standard form.**

Warm Up Write each function in standard form. Evaluate b2 – 4ac for the given values of the valuables. 1. f(x) = (x – 4)2 + 3 2. g(x) = 2(x + 6)2 – 11 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

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**Objectives Solve quadratic equations using the Quadratic Formula.**

Classify roots using the discriminant.

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Vocabulary discriminant

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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.

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**To subtract fractions, you need a common denominator.**

Remember!

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**The symmetry of a quadratic function is evident in the last step,**

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.

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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.

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**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.

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Example 1 Continued Check Solve by completing the square.

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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.

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**Check It Out! Example 1a Continued**

Check Solve by completing the square. x2 + 3x – 7 = 0 x2 + 3x = 7

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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.

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**Check It Out! Example 1b Continued**

Check Solve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 16 = – (x + 4)2 = 6

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**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.

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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.

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

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

Caution!

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**Example 3A: Analyzing Quadratic Equations by Using the Discriminant**

Find the type and number of solutions for the equation. x = 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.

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**Example 3B: Analyzing Quadratic Equations by Using the Discriminant**

Find the type and number of solutions for the equation. x = 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.

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**Example 3C: Analyzing Quadratic Equations by Using the Discriminant**

Find the type and number of solutions for the equation. x = 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.

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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.

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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.

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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) = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

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**The graph shows related functions**

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.

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**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 = –16t t 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?

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Example 4 Continued Step 1 Use the first equation to determine how 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 = –16t t + 6.5 0 = –16t t + 6.5 Set y equal to 0. Use the Quadratic Formula. Substitute –16 for a, 24.6 for b, and 6.5 for c.

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

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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.

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

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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 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 to y = –16t2 –2t To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?

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**Check It Out! Example 4 Continued**

Step 1 Use the equation to determine how 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.

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

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**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 x ≈ 91(4.937) Substitute for t. x ≈ 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.

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**Check It Out! Example 4 Continued**

Check Use substitution to check that the water hits the ground after about seconds. y = –16t2 – 2t + 400 y ≈ –16(4.937)2 – 2(4.937) + 400 y ≈ – – y ≈ 0.143 The height is approximately equal to 0 when t =

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**Properties of Solving Quadratic Equations**

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**Properties of Solving Quadratic Equations**

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

Helpful Hint

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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

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Lesson Quiz: Part II 5. A pebble is tossed from the top of a cliff. The pebble’s height is given by y(t) = –16t , 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

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