Learning Objectives
In this section, you will:
- Solve equations involving rational exponents.
- Solve equations using factoring.
- Solve radical equations.
- 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, 16121612 is another way of writing 16−−√;16; 813813 is another way of writing 8–√3.83. 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, 23(32)=1,23(32)=1, 3(13)=1,3(13)=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:
amn=(a1n)m=(am)1n=am−−−√n=(a−−√n)m𝑎𝑚𝑛=(𝑎1𝑛)𝑚=(𝑎𝑚)1𝑛=𝑎𝑚𝑛=(𝑎𝑛)𝑚
EXAMPLE 1
Evaluating a Number Raised to a Rational Exponent
Evaluate 823.823.
TRY IT #1
Evaluate 64−13.64−13.
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: x54=32.𝑥54=32.
TRY IT #2
Solve the equation x32=125.𝑥32=125.
EXAMPLE 3
Solving an Equation Involving Rational Exponents and Factoring
Solve 3×34=x12.3𝑥34=𝑥12.
TRY IT #3
Solve: (x+5)32=8.(𝑥+5)32=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
anxn+an−1xn−1+⋅⋅⋅+a2x2+a1x+a0𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+⋅⋅⋅+𝑎2𝑥2+𝑎1𝑥+𝑎0
where n is a positive integer and an,…,a0𝑎𝑛,…,𝑎0 are real numbers and an≠0.𝑎𝑛≠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: 5×4=80×2.5𝑥4=80𝑥2.
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: 12×4=3×2.12𝑥4=3𝑥2.
EXAMPLE 5
Solve a Polynomial by Grouping
Solve a polynomial by grouping: x3+x2−9x−9=0.𝑥3+𝑥2−9𝑥−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.
Solving Radical Equations
Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as
3x+18−−−−−−√x+3−−−−√x+5−−−−√−x−3−−−−√===xx−323𝑥+18=𝑥𝑥+3=𝑥−3𝑥+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.
RADICAL EQUATIONS
An equation containing terms with a variable in the radicand is called a radical equation.
HOW TO
Given a radical equation, solve it.
- Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
- 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.
- Solve the remaining equation.
- If a radical term still remains, repeat steps 1–2.
- 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−1𝑥+3=3𝑥−1
EXAMPLE 7
Solving a Radical Equation Containing Two Radicals
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 88 or −8.−8. This leads to two different equations we can solve independently.
2x−62xx===8147or2x−62xx===−8−2−12𝑥−6=8or2𝑥−6=−82𝑥=142𝑥=−2𝑥=7𝑥=−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:
If x≥0,then |x|=x.If x<0,then |x|=−x.If 𝑥≥0,then |𝑥|=𝑥.If 𝑥<0,then |𝑥|=−𝑥.
For real numbers A𝐴 and B,𝐵, an equation of the form |A|=B,|𝐴|=𝐵, with B≥0,𝐵≥0, will have solutions when A=B𝐴=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:
If c<0,|ax+b|=chas no solution.If c=0,|ax+b|=chas one solution.If c>0,|ax+b|=chas two solutions.If 𝑐<0,|𝑎𝑥+𝑏|=𝑐has no solution.If 𝑐=0,|𝑎𝑥+𝑏|=𝑐has one solution.If 𝑐>0,|𝑎𝑥+𝑏|=𝑐has two solutions.
HOW TO
Given an absolute value equation, solve it.
- Isolate the absolute value expression on one side of the equal sign.
- If c>0,𝑐>0, write and solve two equations: ax+b=c𝑎𝑥+𝑏=𝑐 and ax+b=−c.𝑎𝑥+𝑏=−𝑐.
EXAMPLE 8
Solving Absolute Value Equations
Solve the following absolute value equations:
- ⓐ |6x+4|=8|6𝑥+4|=8
- ⓑ |3x+4|=−9|3𝑥+4|=−9
- ⓒ |3x−5|−4=6|3𝑥−5|−4=6
- ⓓ |−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.
Solving Equations in Quadratic Form
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 x4−5×2+4=0,×6+7×3−8=0,𝑥4−5𝑥2+4=0,𝑥6+7𝑥3−8=0, and x23+4×13+2=0.𝑥23+4𝑥13+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.
QUADRATIC FORM
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.
- Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
- If it is, substitute a variable, such as u, for the variable portion of the middle term.
- Rewrite the equation so that it takes on the standard form of a quadratic.
- Solve using one of the usual methods for solving a quadratic.
- Replace the substitution variable with the original term.
- Solve the remaining equation.
EXAMPLE 9
Solving a Fourth-degree Equation in Quadratic Form
Solve this fourth-degree equation: 3×4−2×2−1=0.3𝑥4−2𝑥2−1=0.
TRY IT #8
Solve using substitution: x4−8×2−9=0.𝑥4−8𝑥2−9=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
Solving a Rational Equation Leading to a Quadratic
Solve the following rational equation: −4xx−1+4x+1=−8×2−1.−4𝑥𝑥−1+4𝑥+1=−8𝑥2−1.
TRY IT #10
Solve 3x+2x−2+1x=−2×2−2x.
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