Quadratic formula

Quadratic formula

Warm UpWrite 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

f(x) = x2 – 8x + 19 g(x) = 2x2 + 24x + 61

4. a = 1, b = 3, c = –33. a = 2, b = 7, c = 59 21

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

Objectives

discriminantVocabulary

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.

To subtract fractions, you need a common denominator.

Remember!

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.

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.

Find the zeros of f(x)= 2x2 – 16x + 27 using the Quadratic Formula.

Example 1: Quadratic Functions with Real Zeros

Set f(x) = 0.Write the Quadratic Formula.

Substitute 2 for a, –16 for b, and 27 for c.

Simplify.

Write in simplest form.

2x2 – 16x + 27 = 0

Check Solve by completing the square.

Example 1 Continued

Find the zeros of f(x) = x2 + 3x – 7 using the Quadratic Formula.

Set f(x) = 0.

Write the Quadratic Formula.

Substitute 1 for a, 3 for b, and –7 for c.

Simplify.


Check It Out! Example 1a

x2 + 3x – 7 = 0

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

x2 + 3x = 7

Check It Out! Example 1a Continued

Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula.

Set f(x) = 0.



Simplify.


Check It Out! Example 1b

x2 – 8x + 10 = 0

Check Solve by completing the square.x2 – 8x + 10 = 0

x2 – 8x = –10 x2 – 8x + 16 = –10 + 16

(x + 4)2 = 6

Check It Out! Example 1b Continued

Find the zeros of f(x) = 4x2 + 3x + 2 using the Quadratic Formula.

Example 2: Quadratic Functions with Complex Zeros

Set f(x) = 0.


Substitute 4 for a, 3 for b, and 2 for c.

Simplify.

Write in terms of i.

f(x)= 4x2 + 3x + 2

Find the zeros of g(x) = 3x2 – x + 8 using the Quadratic Formula.

Set f(x) = 0



Simplify.

Write in terms of i.

Check It Out! Example 2

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

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

Caution!

Find the type and number of solutions for the equation.

Example 3A: Analyzing Quadratic Equations by Using the Discriminant

x2 + 36 = 12xx2 – 12x + 36 = 0b2 – 4ac(–12)2 – 4(1)(36)144 – 144 = 0

b2 – 4ac = 0

The equation has one distinct real solution.


Example 3B: Analyzing Quadratic Equations by Using the Discriminant

x2 + 40 = 12xx2 – 12x + 40 = 0b2 – 4ac(–12)2 – 4(1)(40)144 – 160 = –16

b2 –4ac < 0

The equation has two distinct nonreal complex solutions.


Example 3C: Analyzing Quadratic Equations by Using the Discriminant

x2 + 30 = 12xx2 – 12x + 30 = 0b2 – 4ac(–12)2 – 4(1)(30)144 – 120 = 24

b2 – 4ac > 0

The equation has two distinct real solutions.

Check It Out! Example 3a Find the type and number of solutions for the equation.

x2 – 4x = –4x2 – 4x + 4 = 0b2 – 4ac(–4)2 – 4(1)(4)16 – 16 = 0

b2 – 4ac = 0

The equation has one distinct real solution.

Check It Out! Example 3b Find the type and number of solutions for the equation.

x2 – 4x = –8x2 – 4x + 8 = 0b2 – 4ac(–4)2 – 4(1)(8)16 – 32 = –16

b2 – 4ac < 0

The equation has two distinct nonreal complex solutions.

Check It Out! Example 3c Find the type and number of solutions for each equation.

x2 – 4x = 2x2 – 4x – 2 = 0b2 – 4ac(–4)2 – 4(1)(–2)16 + 8 = 24

b2 – 4ac > 0

The equation has two distinct real solutions.

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.

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?

Example 4: Sports Application

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 = –16t2 + 24.6t + 6.5

Example 4 Continued

Set y equal to 0.0 = –16t2 + 24.6t + 6.5

Use the Quadratic Formula.

Substitute –16 for a, 24.6 for b, and 6.5 for c.

Example 4 Continued

Simplify.

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

Step 2 Find the horizontal distance that the shot put will have traveled in this time.

x = 29.3t

Example 4 Continued

Substitute 1.77 for t.

Simplify.

x ≈ 29.3(1.77)

x ≈ 51.86x ≈ 52

The shot put will have traveled a horizontal distance of about 52 feet.

Check Use substitution to check that the shot put hits the ground after about 1.77 seconds.

Example 4 Continued

The height is approximately equal to 0 when t = 1.77.

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 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?

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.

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 + 400Set y equal to 0.0 = –16t2 – 2t + 400


Substitute –16 for a, –2 for b, and 400 for c.

Check It Out! Example 4 Continued


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.

Check It Out! Example 4 Continued

Simplify.

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 = 91tSubstitute 4.937 for t.x ≈ 91(4.937)

x ≈ 449.267x ≈ 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.

Check It Out! Example 4 Continued Check Use substitution to check that the water hits

the ground after about 4.937 seconds.

The height is approximately equal to 0 when t = 4.937.

y = –16t2 – 2t + 400 y ≈ –16(4.937)2 – 2(4.937) + 400y ≈ –389.983 – 9.874 + 400y ≈ 0.143

Properties of Solving Quadratic Equations

Properties of Solving Quadratic Equations

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

Helpful Hint

Lesson Quiz: Part IFind 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 + 482 distinct nonreal complex

2 distinct real

Lesson Quiz: Part II5. 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

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Quadratic formula