Copyright © 2010 Pearson Education, Inc.

Quadratic Equations Quadratic Equations and Problem Solvingand Problem Solving

♦ Understand basic concepts about quadratic Understand basic concepts about quadratic equationsequations

♦ Use factoring, the square root property, Use factoring, the square root property, completing the square, and the quadratic completing the square, and the quadratic formula to solve quadratic equationsformula to solve quadratic equations

♦ Understand the discriminantUnderstand the discriminant♦ Solve problems involving quadratic equationsSolve problems involving quadratic equations

3.23.2

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Quadratic Equation

A quadratic equation in one variable is an equation that can be written in the form

ax2 + bx + c

where a, b, and c are constants with a ≠ 0.

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Solving Quadratic EquationsQuadratic equations can have no real solutions, one real solution, or two real solutions.The are four basic symbolic strategies in which quadratic equations can be solved.

• Factoring• Square root property• Completing the square• Quadratic formula

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FactoringFactoring is a common technique used to solve equations. It is based on the zero-product property, which states that if

ab = 0, then a = 0 or b = 0 or both. It is important to remember that this property works only for 0. For example, if ab = 1, then this equation does not imply that either a = 1 or b = 1. For example, a = 1/2 and b = 2 also satisfies ab = 1 and neither a nor b is 1.

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Example 1bSolve the quadratic equation 12t2 = t + 1. Check your results. Check:

Solution

12t 2 t 1

12t 2 t 10

3t 1 4t 1 0

3t 10 or 4t 10

t 1

3 or t

1

4

121

3

2

1

31

4

3

4

3

12 1

4

2

1

41

3

4

3

4

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Example 1bSolution continued

Check:

121

3

2

1

31

4

3

4

3

12 1

4

2

1

41

3

4

3

4

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The Square Root Property

Let k be a nonnegative number. Then the solutions to the equation

x2 = k

are given by x k .

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Example 4

If a metal ball is dropped 100 feet from a water tower, its height h in feet above the ground after t seconds is given by h(t) = 100 – 16t2. Determine how long it takes the ball to hit the ground.

Solution

The ball strikes the ground when the equation 100 – 16t2 = 0 is satisfied.

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Example 4

Solution continued

The ball strikes the ground after 10/4, or 2.5, seconds.

100 16t 2 0

100 16t 2

t 2 100

16

t 100

16

10

4

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Completing the Square

Another technique that can be used to solve a quadratic equation is completing the square. If a quadratic equation is written in the form x2 + kx =d, where k and d are constants, then the equation can be solved using

x2 kx

k

2

2

x k

2

2

.

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Example

Solve 2x2 – 8x = 7.

Solution

Divide each side by 2:

x 2 15

2

x 2 15

2

x2 4x 7

2

x2 4x 4 7

2 4

x 2 2 15

2

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Symbolic, Numerical and Graphical Solutions

Quadratic equations can be solved symbolically, numerically, and graphically. The following example illustrates each technique for the equation x(x – 2) = 3.

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Symbolic Solution

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Numerical Solution

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Graphical Solution

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Quadratic Formula

The solutions to the quadratic equation

ax2 + bx + c = 0, where a ≠ 0, are given by

x

b b2 4ac

2a.

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Example 7Solve the equation 3x2 – 6x + 2 = 0.

Solution

Let a = 3, b = 6, and c = 2.

x b b2 4ac

2a

x 6 6 2 4 3 2

2 3 x

6 12

61

1

343 1

1

33

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The Discriminant

If the quadratic equation ax2 + bx + c = 0 is solved graphically, the parabolay = ax2 + bx + c can intersect the x-axis zero, one, or two times. Each x-intercept is a real solution to the quadratic equation.

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Quadratic Equations and Discriminant

To determine the number of real solutions to ax2 + bx + c = 0 with a ≠ 0, evaluate the discriminant b2 – 4ac.

1. If b2 – 4ac > 0, there are two real solutions.

2. If b2 – 4ac = 0, there is one real solution.

3. If b2 – 4ac < 0, there are no real solutions.

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Example 9

Use the discriminant to find the number of solutions to 9x2 – 12.6x + 4.41 = 0. Then solve the equation by using the quadratic formula. Support your answer graphically.

Solution

Let a = 9, b = –12.6, and c = 4.41

b2 – 4ac = (–12.6)2 – 4(9)(4.41) = 0

Discriminant is 0, there is one solution.

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Example 9

Solution continued

The only solution is 0.7.

x b b2 4ac

2a

x 12.6 0

18x 0.7

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Example 9

Solution continued

The graph suggests there is only one intercept 0.7.

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Problem Solving

Many types of applications involve quadratic equations. To solve these problems, we use the steps for “Solving Application Problems” from Section 2.3 on page 122.

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Example 11

A box is being constructed by cutting2-inch squares from the corners of a rectangular piece of cardboard that is 6 inches longer than it is wide. If the box is to have a volume of 224 cubic inches, find

the dimensionsof the pieceof cardboard.

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Example 11

Solution

Step 1: Let x be the width and x + 6 be the length.

Step 2: Draw a picture.

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Solution continued

Since the height times the width times the length must equal the volume, or 224 cubic inches, the following can be written

2(x – 4)(x + 2) = 224

Step 3: Write the quadratic equation in the form ax2 + bx + c = 0 and factor.

x2 2x 8 112

x2 2x 120 0

x 12 x 10 0

x 12 or x 10

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Solution continued

The dimensions can not be negative, so the width is 12 inches and the length is 6 inches more, or 18 inches.

Step 4: After the 2-inch-square corners are cut out, the dimensions of the bottom of the box are 12 – 4 = 8 inches by 18 – 4 = 14 inches. The volume of the box is then 2•8•14 = 224 cubic inches, which checks.