1.4 QUADRATIC EQUATIONS AND APPLICATIONS Copyright © Cengage Learning. All rights reserved.

1.4 QUADRATIC EQUATIONS AND APPLICATIONS

Copyright © Cengage Learning. All rights reserved.

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• Solve quadratic equations by factoring.

• Solve quadratic equations by extracting square roots.

• Solve quadratic equations by completing the square.

• Use the Quadratic Formula to solve quadratic equations.

• Use quadratic equations to model and solve real-life problems.

What You Should Learn

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Factoring

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Example 1(a) – Solving a Quadratic Equation by Factoring

2x2 + 9x + 7 = 3

2x2 + 9x + 4 = 0

(2x + 1)(x + 4) = 0

2x + 1 = 0 x =

x + 4 = 0 x = –4

The solutions are x = and x = –4. Check these in the original equation.

Original equation

Write in general form.

Factor.

Set 1st factor equal to 0.

Set 2nd factor equal to 0.

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Example 1(b) – Solving a Quadratic Equation by Factoring

6x2 – 3x = 0

3x(2x – 1) = 0

3x = 0 x = 0

2x – 1 = 0 x =

The solutions are x = 0 and x = . Check these in the original equation.

Original equation

Set 1st factor equal to 0.

Factor.

Set 2nd factor equal to 0.

cont’d

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Extracting Square Roots

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Example 2 – Extracting Square RootsSolve each equation by extracting square roots.

a. 4x2 = 12 b. (x – 3)2 = 7

Solution:

a. 4x2 = 12

x2 = 3

x =

When you take the square root of a variable expression,

you must account for both positive and negative solutions.

Write original equation.

Divide each side by 4.

Extract square roots.

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Example 2 – Solution

So, the solutions are x = and x = – . Check these in the original equation.

b. (x – 3)2 = 7

x – 3 =

x = 3

The solutions are x = 3 . Check these in the original equation.

Write original equation.

Extract square roots.

Add 3 to each side.

cont’d

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Completing the Square

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Example 3 – Completing the Square: Leading Coefficient Is 1

Solve x2 + 2x – 6 = 0 by completing the square.

Solution:

x2 + 2x – 6 = 0

x2 + 2x = 6

x2 + 2x + 12 = 6 + 12

Write original equation.

Add 6 to each side.

Add 12 to each side.

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Example 3 – Solution

(x + 1)2 = 7

x + 1 =

x = –1

The solutions are x = –1 . Check these in the original equation as follows.

Simplify.

Take square root of each side.

Subtract 1 from each side.

cont’d

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Example 3 – Solution

Check:

x2 + 2x – 6 = 0

(–1 + )2 + 2 (–1 + ) – 6 ≟ 0

8 – 2 – 2 + 2 – 6 ≟ 0

8 – 2 – 6 = 0

Check the second solution in the original equation.

Write original equation.

Substitute –1 + for x.

Multiply.

Solution checks.

cont’d

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The Quadratic Formula

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The Quadratic Formula

The Quadratic Formula is one of the most important formulas in algebra. You should learn the verbal statement of the Quadratic Formula:

“Negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.”

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The Quadratic Formula

In the Quadratic Formula, the quantity under the radical sign, b2 – 4ac, is called the discriminant of the quadratic expression ax2 + bx + c. It can be used to determine the nature of the solutions of a quadratic equation.

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Example 6 – The Quadratic Formula: Two Distinct Solutions

Use the Quadratic Formula to solve x2 + 3x = 9.

Solution:

The general form of the equation is x2 + 3x – 9 = 0. The discriminant is b2 – 4ac = 9 + 36 = 45, which is positive. So, the equation has two real solutions.

You can solve the equation as follows.

x2 + 3x – 9 = 0 Write in general form.

Quadratic Formula

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Example 6 – Solution

The two solutions are:

Check these in the original equation.

Substitute a = 1, b = 3,and c = –9.

Simplify.

Simplify.

cont’d

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Applications

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Example 7 – Finding the Dimensions of a Room

A bedroom is 3 feet longer than it is wide (see Figure 1.20) and has an area of 154 square feet. Find the dimensions of the room.

Figure 1.20

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Example 7 – Solution

Verbal

Model:

Labels: Width of room = w (feet)

Length of room = w + 3 (feet)

Area of room = 154 (square feet)

Equation: w(w + 3) = 154

w2 + 3w – 154 = 0

(w – 11)(w + 14) = 0

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Example 7 – Solution

w – 11 = 0 w = 11

w + 14 = 0 w = –14

Choosing the positive value, you find that the width is 11 feet and the length is w + 3, or 14 feet.

You can check this solution by observing that the length is 3 feet longer than the width and that the product of the length and width is 154 square feet.

cont’d

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Applications

Another common application of quadratic equations involves an object that is falling (or projected into the air).

The general equation that gives the height of such an object is called a position equation, and on Earth’s surface it has the form

s = –16t2 + v0t + s0.

In this equation, s represents the height of the object (in feet), v0 represents the initial velocity of the object (in feet per second), s0 represents the initial height of the object (in feet), and t represents the time (in seconds).

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Applications

A third type of application of a quadratic equation is one in which a quantity is changing over time t according to a quadratic model.

A fourth type of application that often involves a quadratic equation is one dealing with the hypotenuse of a right triangle.

In these types of applications, the Pythagorean Theorem is often used.

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Applications

The Pythagorean Theorem states that

a2 + b2 = c2

where a and b are the legs of a right triangle and c is the hypotenuse.

Pythagorean Theorem