1. Copyright © Cengage Learning. All rights reserved.1
Equations, Inequalities, and
Mathematical Modeling
Copyright © Cengage Learning. All rights reserved.

2. Other Types of Equations
1.6
Other Types of Equations
Copyright © Cengage Learning. All rights reserved.

3. Objectives Solve polynomial equations of degree three or greater.
Solve radical equations.
Solve rational equations and absolute value equations.
Use nonlinear and nonquadratic models to solve real-life problems.

4. Polynomial Equations

5. 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.

6. Polynomial Equations
Example 1 shows how to use factoring to solve a polynomial equation, which is an equation that can be written in the general form anxn + an – 1 xn – a2x2 + a1x + a0 = 0.

7. 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. Then factor the other side, 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.
Write in factored form.
Set 1st factor equal to 0.

8. Example 1 – Solution
cont’d
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.
Set 2nd factor equal to 0.
Set 3rd factor equal to 0.
0 checks.
–4 checks.
4 checks.

9. 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.

10. Polynomial Equations
Occasionally, mathematical models involve equations that are of quadratic type. In general, an equation is of quadratic type when it can be written in the form au2 + bu + c = 0 where a  0 and u is an algebraic expression.

11. Radical Equations

12. Radical Equations
A radical equation is an equation that involves one or more radical expressions.

13. Example 4 – Solving Radical Equations
a. 2x + 7 = x2 + 4x = 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.

14. Example 4 – Solving Radical Equations
cont’d
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

15. Example 4 – Solving Radical Equations
cont’d
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.

16. Rational Equations and Absolute Value Equations

17. Rational Equations and Absolute Value Equations
Recall that the first step is to multiply each term of the equation by the least common denominator (LCD).

18. Example 6 – Solving a Rational Equation
Solve Solution: For this equation, the LCD 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.

19. Example 6 – Solution
cont’d
2(x – 2) = 3x – x(x – 2), x  0, 2 2x – 4 = –x2 + 5x x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x – 4 = 0 x = 4 x + 1 = 0 x = –1 Both x = 4 and x = –1 are possible solutions.
Simplify.
Simplify.
Write in general form.
Factor.
Set 1st factor equal to 0.
Set 2nd factor equal to 0.

20. Example 6 – Solution
cont’d
Because multiplying each side of an equation by a variable expression can introduce extraneous solutions, it is important to check your solutions. Check x = 4 Check x = –1

21. Example 6 – Solution
cont’d
So, the solutions are x = 4 and x = –1.

22. Rational Equations and Absolute Value Equations
An absolute value equation is an equation that involves one or more absolute value expressions. To solve an absolute value equation, remember that the expression inside the absolute value bars 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 and x = –1.

23. Applications

24. Applications
It would be virtually impossible to categorize all of the many different types of applications that involve nonlinear and nonquadratic models. However, you will see a variety of applications in the examples and exercises.

25. Example 8 – Reduced Rates
A ski club charters a bus for a ski trip at a cost of $480. To lower the bus fare per skier, the club invites nonmembers to go along. After 5 nonmembers join the trip, the fare per skier decreases by $4.80. How many club members are going on the trip? Solution:

26. Example 8 – Solution
cont’d
Labels: Cost of trip = 480 (dollars) Number of ski club members = x (people) Number of skiers = x + 5 (people) Original cost per member = (dollars per person) Cost per skier = (dollars per person) Equation:
Rewrite first factor.

27. Example 8 – Solution
cont’d
(480 – 4.8x)(x + 5) = 480x, x  0 480x – 4.8x2 – 24x = 480x –4.8x2 – 24x = 0 x2 + 5x – 500 = 0 (x + 25)(x – 20) = 0 x + 25 = 0 x = –25 x – 20 = 0 x = 20
Multiply each side by x.
Multiply.
Subtract 480x from each side.
Divide each side by –4.8.
Factor.
Set 1st factor equal to 0.
Set 2nd factor equal to 0.

28. Example 8 – Solution
cont’d
Only the positive value of x makes sense in the context of the problem, so you can conclude that 20 ski club members are going on the trip. Check this in the original statement of the problem, as follows. Check: (24 – 4.80)25 ≟ = 480
Substitute 20 for x.
Simplify.
20 checks.