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P. 1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.
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P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

Jan 02, 2016

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Page 1: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

P.1 OTHER TYPES OF EQUATIONS

Copyright © Cengage Learning. All rights reserved.

Page 2: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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• Solve polynomial equations of degree three or greater.

• Solve equations involving radicals.

• Solve equations involving fractions or absolute values.

• Use polynomial equations and equations involving radicals to model and solve real-life problems.

What You Should Learn

Page 3: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Polynomial Equations

Page 4: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Polynomial Equations

In this section you will extend the techniques for solving equations to nonlinear and nonquadratic equations.

At this point in the text, you have only four basic methods for solving nonlinear equations—factoring, extracting square roots, completing the square, and the Quadratic Formula.

So the main goal of this section is to learn to rewrite nonlinear equations in a form to which you can apply one of these methods.

Page 5: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Example 1 – Solving a Polynomial Equation by Factoring

Solve 3x4 = 48x2.

Solution:

First write the polynomial equation in general form with zero on one side, factor the other side, and then set each factor equal to zero and solve.

3x4 = 48x2

3x4 – 48x2 = 0

3x2(x2 – 16) = 0

3x2(x + 4)(x – 4) = 0

3x2 = 0 x = 0

Write original equation.

Write in general form.

Factor out common factor.

Set 1st factor equal to 0.

Write in factored form.

Page 6: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Example 1 – Solution

x + 4 = 0 x = –4

x – 4 = 0 x = 4

You can check these solutions by substituting in the original equation, as follows.

Check

3(0)4 = 48(0)2

3(–4)4 = 48(–4)2

3(4)4 = 48(4)2

So, you can conclude that the solutions are x = 0, x = –4, and x = 4.

cont’d

Set 2nd factor equal to 0.

Set 3rd factor equal to 0.

0 checks.

–4 checks.

4 checks.

Page 7: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Polynomial Equations

A common mistake that is made in solving an equation like that in Example 1 is to divide each side of the equation by the variable factor x2.

This loses the solution x = 0. When solving an equation, always write the equation in general form, then factor the equation and set each factor equal to zero.

Do not divide each side of an equation by a variable factor in an attempt to simplify the equation.

Page 8: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Solving a Polynomial Equation: Factoring By Grouping

Solve x3 – 3x2 – 3x+ 9 = 0.

Page 9: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Equations Involving Radicals

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Equations Involving Radicals

Operations such as squaring each side of an equation, raising each side of an equation to a rational power, and multiplying each side of an equation by a variable quantity all can introduce extraneous solutions.

So, when you use any of these operations, checking your solutions is crucial.

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Example 4 – Solving Equations Involving Radicals

a.

2x + 7 = x2 + 4x + 4

0 = x2 + 2x – 3

0 = (x + 3)(x – 1)

x + 3 = 0 x = –3

Original equation

Isolate radical.

Square each side.

Write in general form.

Factor.

Set 1st factor equal to 0.

Page 12: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Example 4 – Solving Equations Involving Radicals

x – 1 = 0 x = 1

By checking these values, you can determine that the only solution is x = 1.

b.

Set 2nd factor equal to 0.

Original equation

Isolate

Square each side.

Combine like terms.

Isolate

Page 13: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Example 4 – Solving Equations Involving Radicals

x2 – 6x + 9 = 4(x – 3)

x2 – 10x + 21 = 0

(x – 3)(x – 7) = 0

x – 3 = 0 x = 3

x – 7 = 0 x = 7

The solutions are x = 3 and x = 7. Check these in the original equation.

Square each side.

Write in general form.

Factor.

Set 1st factor equal to 0.

Set 2nd factor equal to 0.

Page 14: P.1 OTHER TYPES OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

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Solving an Equation Involving a Rational Exponent

Solve (x – 4)2/3 = 25.

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Equations with Fractions or Absolute Values

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Equations with Fractions or Absolute Values

To solve an equation involving fractions, multiply each side of the equation by the least common denominator (LCD) of all terms in the equation.

This procedure will “clear the equation of fractions.” For instance, in the equation

you can multiply each side of the equation by x(x2 + 1). Try doing this and solve the resulting equation. You should obtain one solution: x = 1.

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Example 6 – Solving an Equation Involving Fractions

Solve

Solution:

For this equation, the least common denominator of the three terms is x(x – 2), so you begin by multiplying each term of the equation by this expression.

Write original equation.

Multiply each term by the LCD.

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Example 6 – Solution

2(x – 2) = 3x – x(x – 2)

2x – 4 = –x2 + 5x

x2 – 3x – 4 = 0

(x – 4)(x + 1) = 0

x – 4 = 0 x = 4

x + 1 = 0 x = –1

Simplify.

Simplify.

Write in general form.

Factor.

Set 1st factor equal to 0.

Set 2nd factor equal to 0.

cont’d

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Example 6 – Solution

Check x = 4 Check x = –1

So, the solutions are x = 4 and x = –1.

cont’d

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Equations with Fractions or Absolute Values

To solve an equation involving an absolute value, remember that the expression inside the absolute value signs can be positive or negative.

This results in two separate equations, each of which must be solved. For instance, the equation

| x – 2 | = 3

results in the two equations x – 2 = 3 and –(x – 2) = 3, which implies that the equation has two solutions: x = 5, x = –1.

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Solving an Equation Involving Absolute Value

Solve 6432 xxx