College Algebra

2.6Other Types of Equations

College Algebra2.6 Other Types of Equations

Learning Objectives

In this section you will:
• Solve equations involving rational exponents.
• Solve equations using factoring.
• Solve absolute value equations.
• Solve other types of equations.

We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.

Solving Equations Involving Rational Exponents

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example,$16 1 2 16 1 2$is another way of writing$16 ; 16 ;$$8 1 3 8 1 3$is another way of writing$​ 8 3 . ​ 8 3 .$The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example,$2 3 ( 3 2 )=1, 2 3 ( 3 2 )=1,$$3( 1 3 )=1, 3( 1 3 )=1,$and so on.

Rational Exponents

A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:

$a m n = ( a 1 n ) m = ( a m ) 1 n = a m n = ( a n ) m a m n = ( a 1 n ) m = ( a m ) 1 n = a m n = ( a n ) m$

Example 1

Evaluating a Number Raised to a Rational Exponent

Evaluate$8 2 3 . 8 2 3 .$

Try It #1

Evaluate$64 − 1 3 . 64 − 1 3 .$

Example 2

Solve the Equation Including a Variable Raised to a Rational Exponent

Solve the equation in which a variable is raised to a rational exponent:$x 5 4 =32. x 5 4 =32.$

Try It #2

Solve the equation$x 3 2 =125. x 3 2 =125.$

Example 3

Solving an Equation Involving Rational Exponents and Factoring

Solve$3 x 3 4 = x 1 2 . 3 x 3 4 = x 1 2 .$

Try It #3

Solve:$( x+5 ) 3 2 =8. ( x+5 ) 3 2 =8.$

Solving Equations Using Factoring

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

Polynomial Equations

A polynomial of degree n is an expression of the type

$a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 2 x 2 + a 1 x+ a 0 a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 2 x 2 + a 1 x+ a 0$

where n is a positive integer and$a n ,…, a 0 a n ,…, a 0$are real numbers and$a n ≠0. a n ≠0.$

Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n.

Example 4

Solving a Polynomial by Factoring

Solve the polynomial by factoring:$5 x 4 =80 x 2 . 5 x 4 =80 x 2 .$

Analysis

We can see the solutions on the graph in Figure 1. The x-coordinates of the points where the graph crosses the x-axis are the solutions—the x-intercepts. Notice on the graph that at the solution$0, 0,$the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions.

Figure 1
Try It #4

Solve by factoring:$12 x 4 =3 x 2 . 12 x 4 =3 x 2 .$

Example 5

Solve a Polynomial by Grouping

Solve a polynomial by grouping:$x 3 + x 2 −9x−9=0. x 3 + x 2 −9x−9=0.$

Analysis

We looked at solving quadratic equations by factoring when the leading coefficient is 1. When the leading coefficient is not 1, we solved by grouping. Grouping requires four terms, which we obtained by splitting the linear term of quadratic equations. We can also use grouping for some polynomials of degree higher than 2, as we saw here, since there were already four terms.

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

$3x+18 = x x+3 = x−3 x+5 − x−3 = 2 3x+18 = x x+3 = x−3 x+5 − x−3 = 2$

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

An equation containing terms with a variable in the radicand is called a radical equation.

How To

Given a radical equation, solve it.

1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol.
3. Solve the remaining equation.
4. If a radical term still remains, repeat steps 1–2.
5. Confirm solutions by substituting them into the original equation.

Example 6

Solving an Equation with One Radical

Solve$15−2x =x. 15−2x =x.$

Try It #5

Solve the radical equation:$x+3 =3x−1 x+3 =3x−1$

Example 7

Solve$2x+3 + x−2 =4. 2x+3 + x−2 =4.$

Try It #6

Solve the equation with two radicals:$3x+7 + x+2 =1. 3x+7 + x+2 =1.$

Solving an Absolute Value Equation

Next, we will learn how to solve an absolute value equation. To solve an equation such as$| 2x−6 |=8, | 2x−6 |=8,$we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is$8 8$or$−8. −8.$This leads to two different equations we can solve independently.

$2x−6 = 8 or 2x−6 = −8 2x = 14 2x = −2 x = 7 x = −1 2x−6 = 8 or 2x−6 = −8 2x = 14 2x = −2 x = 7 x = −1$

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

Absolute Value Equations

The absolute value of x is written as$| x |. | x |.$It has the following properties:

For real numbers$A A$and$B, B,$an equation of the form$| A |=B, | A |=B,$with$B≥0, B≥0,$will have solutions when$A=B A=B$or$A=−B. A=−B.$If$B<0, B<0,$the equation$| A |=B | A |=B$has no solution.

An absolute value equation in the form$| ax+b |=c | ax+b |=c$has the following properties:

How To

Given an absolute value equation, solve it.

1. Isolate the absolute value expression on one side of the equal sign.
2. If$c>0, c>0,$write and solve two equations:$ax+b=c ax+b=c$and$ax+b=−c. ax+b=−c.$

Example 8

Solving Absolute Value Equations

Solve the following absolute value equations:

• (a) $| 6x+4 |=8 | 6x+4 |=8$
• (b) $| 3x+4 |=−9 | 3x+4 |=−9$
• (c) $| 3x−5 |−4=6 | 3x−5 |−4=6$
• (d) $| −5x+10 |=0 | −5x+10 |=0$
Try It #7

Solve the absolute value equation:$| 1−4x |+8=13. |1−4x|+8=13.$

Solving Other Types of Equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include$x 4 −5 x 2 +4=0, x 6 +7 x 3 −8=0, x 4 −5 x 2 +4=0, x 6 +7 x 3 −8=0,$and$x 2 3 +4 x 1 3 +2=0. x 2 3 +4 x 1 3 +2=0.$In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form, which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.

How To

Given an equation quadratic in form, solve it.

1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
2. If it is, substitute a variable, such as u, for the variable portion of the middle term.
3. Rewrite the equation so that it takes on the standard form of a quadratic.
4. Solve using one of the usual methods for solving a quadratic.
5. Replace the substitution variable with the original term.
6. Solve the remaining equation.

Example 9

Solving a Fourth-degree Equation in Quadratic Form

Solve this fourth-degree equation:$3 x 4 −2 x 2 −1=0. 3 x 4 −2 x 2 −1=0.$

Try It #8

Solve using substitution:$x 4 −8 x 2 −9=0. x 4 −8 x 2 −9=0.$

Example 10

Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form:$( x+2 ) 2 +11( x+2 )−12=0. ( x+2 ) 2 +11( x+2 )−12=0.$

Try It #9

Solve:$( x−5 ) 2 −4( x−5 )−21=0. ( x−5 ) 2 −4( x−5 )−21=0.$

Solving Rational Equations Resulting in a Quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Example 11

Solve the following rational equation:$−4x x−1 + 4 x+1 = −8 x 2 −1 . −4x x−1 + 4 x+1 = −8 x 2 −1 .$

Try It #10

Solve$3x+2 x−2 + 1 x = −2 x 2 −2x . 3x+2 x−2 + 1 x = −2 x 2 −2x .$

Media

Access these online resources for additional instruction and practice with different types of equations.

2.6 Section Exercises

Verbal

1.

In a radical equation, what does it mean if a number is an extraneous solution?

2.

Explain why possible solutions must be checked in radical equations.

3.

Your friend tries to calculate the value$− 9 3 2 − 9 3 2$and keeps getting an ERROR message. What mistake is he or she probably making?

4.

Explain why$| 2x+5 |=−7 | 2x+5 |=−7$has no solutions.

5.

Explain how to change a rational exponent into the correct radical expression.

Algebraic

For the following exercises, solve the rational exponent equation. Use factoring where necessary.

6.

$x 2 3 =16 x 2 3 =16$

7.

$x 3 4 =27 x 3 4 =27$

8.

$2 x 1 2 − x 1 4 =0 2 x 1 2 − x 1 4 =0$

9.

$( x−1 ) 3 4 =8 ( x−1 ) 3 4 =8$

10.

$( x+1 ) 2 3 =4 ( x+1 ) 2 3 =4$

11.

$x 2 3 −5 x 1 3 +6=0 x 2 3 −5 x 1 3 +6=0$

12.

$x 7 3 −3 x 4 3 −4 x 1 3 =0 x 7 3 −3 x 4 3 −4 x 1 3 =0$

For the following exercises, solve the following polynomial equations by grouping and factoring.

13.

$x 3 +2 x 2 −x−2=0 x 3 +2 x 2 −x−2=0$

14.

$3 x 3 −6 x 2 −27x+54=0 3 x 3 −6 x 2 −27x+54=0$

15.

$4 y 3 −9y=0 4 y 3 −9y=0$

16.

$x 3 +3 x 2 −25x−75=0 x 3 +3 x 2 −25x−75=0$

17.

$m 3 + m 2 −m−1=0 m 3 + m 2 −m−1=0$

18.

$2 x 5 −14 x 3 =0 2 x 5 −14 x 3 =0$

19.

$5 x 3 +45x=2 x 2 +18 5 x 3 +45x=2 x 2 +18$

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

20.

$3x−1 −2=0 3x−1 −2=0$

21.

$x−7 =5 x−7 =5$

22.

$x−1 =x−7 x−1 =x−7$

23.

$3t+5 =7 3t+5 =7$

24.

$t+1 +9=7 t+1 +9=7$

25.

$12−x =x 12−x =x$

26.

$2x+3 − x+2 =2 2x+3 − x+2 =2$

27.

$3x+7 + x+2 =1 3x+7 + x+2 =1$

28.

$2x+3 − x+1 =1 2x+3 − x+1 =1$

For the following exercises, solve the equation involving absolute value.

29.

$| 3x−4 |=8 | 3x−4 |=8$

30.

$| 2x−3 |=−2 | 2x−3 |=−2$

31.

$| 1−4x |−1=5 | 1−4x |−1=5$

32.

$| 4x+1 |−3=6 | 4x+1 |−3=6$

33.

$| 2x−1 |−7=−2 | 2x−1 |−7=−2$

34.

$| 2x+1 |−2=−3 | 2x+1 |−2=−3$

35.

$| x+5 |=0 | x+5 |=0$

36.

$−| 2x+1 |=−3 −| 2x+1 |=−3$

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

37.

$x 4 −10 x 2 +9=0 x 4 −10 x 2 +9=0$

38.

$4 ( t−1 ) 2 −9( t−1 )=−2 4 ( t−1 ) 2 −9( t−1 )=−2$

39.

$( x 2 −1 ) 2 +( x 2 −1 )−12=0 ( x 2 −1 ) 2 +( x 2 −1 )−12=0$

40.

$( x+1 ) 2 −8( x+1 )−9=0 ( x+1 ) 2 −8( x+1 )−9=0$

41.

$( x−3 ) 2 −4=0 ( x−3 ) 2 −4=0$

Extensions

For the following exercises, solve for the unknown variable.

42.

$x −2 − x −1 −12=0 x −2 − x −1 −12=0$

43.

$| x | 2 =x | x | 2 =x$

44.

$t 10 −2 t 5 +1=0 t 10 −2 t 5 +1=0$

45.

$| x 2 +2x−36 |=12 | x 2 +2x−36 |=12$

Real-World Applications

For the following exercises, use the model for the period of a pendulum,$T, T,$such that$T=2π L g , T=2π L g ,$where the length of the pendulum is L and the acceleration due to gravity is$g. g.$

46.

If the acceleration due to gravity is 9.8 m/s2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).

47.

If the gravity is 32 ft/s2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.

For the following exercises, use a model for body surface area, BSA, such that$BSA= wh 3600 , BSA= wh 3600 ,$where w = weight in kg and h = height in cm.

48.

Find the height of a 72-kg female to the nearest cm whose$BSA=1.8. BSA=1.8.$

49.

Find the weight of a 177-cm male to the nearest kg whose$BSA=2.1. BSA=2.1.$