1. 1 Equations, Inequalities, and Mathematical Modeling
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2. Copyright © Cengage Learning. All rights reserved.
1.6
OTHER TYPES OF EQUATIONS
Copyright © Cengage Learning. All rights reserved.

3. What You Should Learn
• 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.

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, 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 = 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 x + 4 = 0 x = –4 x – 4 = 0 x = 4
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 if it can be written in the form
au2 + bu + c = 0
where a  0 and u is an algebraic expression.

11. Equations Involving Radicals

12. 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.

13. Example 4 – Solving Equations Involving Radicals
2x + 7 = x2 + 4x + 4
0 = x2 + 2x – 3
0 = (x + 3)(x – 1)
x + 3 = x = –3
Original equation
Isolate radical.
Square each side.
Write in general form.
Factor.
Set 1st factor equal to 0.

14. Example 4 – Solving Equations Involving Radicals
x – 1 = 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 Equations Involving Radicals
x2 – 6x + 9 = 4(x – 3)
x2 – 10x + 21 = 0
(x – 3)(x – 7) = 0
x – 3 = x = 3
x – 7 = 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. Equations with Fractions or Absolute Values

17. 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.

18. 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.

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

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

21. 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.

22. Applications

23. Applications
It would be impossible to categorize the many different types of applications that involve nonlinear and nonquadratic models.
However, from the few examples and exercises that are given, you will gain some appreciation for the variety of applications that can occur.

24. Example 8 – Reduced Rates
A ski club chartered a bus for a ski trip at a cost of $480. In an attempt to lower the bus fare per skier, the club invited nonmembers to go along. After five nonmembers joined the trip, the fare per skier decreased by $4.80. How many club members are going on the trip?
Solution:
Begin the solution by creating a verbal model and assigning labels.

25. Example 8 – Solution Labels: Cost of trip = 480 (dollars)
cont’d
Labels: Cost of trip = (dollars)
Number of ski club members = x (people)
Number of skiers = x (people)
Original cost per member = (dollars per person)
Cost per skier = (dollars per person)
Equation:
Write as a fraction.

26. Example 8 – Solution Equation: (480 – 4.8x)(x + 5) = 480x
cont’d
Equation: (480 – 4.8x)(x + 5) = 480x
480x – 4.8x2 – 24x = 480x
–4.8x2 – 24x = 0
x2 + 5x – 500 = 0
(x + 25)(x – 20) = 0
x + 25 = x = –25
x – 20 = x = 20
Multiply each side by x.
Multiply.
Subtract 480x from each side.
Divide each side by –4.8.
Factor.

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

28. Applications
Interest in a savings account is calculated by one of three basic methods: simple interest, interest compounded n times per year, and interest compounded continuously.
The next example uses the formula for interest that is compounded n times per year.
In this formula, A is the balance in the account, P is the principal (or original deposit), r is the annual interest rate (in decimal form), n is the number of compoundings per year, and t is the time in years.