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Guide allo studio > College Algebra

Equation-Solving Techniques

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.

Equations With Radicals and Rational Exponents

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as
3x+18=xx+3=x−3x+5−x−3=2\begin{array}{l}\sqrt{3x+18}\hfill&=x\hfill \\ \sqrt{x+3}\hfill&=x - 3\hfill \\ \sqrt{x+5}-\sqrt{x - 3}\hfill&=2\hfill \end{array}
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.

A General Note: 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.

  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: Solving an Equation with One Radical

Solve 15−2x=x\sqrt{15 - 2x}=x.

Answer: The radical is already isolated on the left side of the equal side, so proceed to square both sides.

15−2x=x(15−2x)2=(x)215−2x=x2\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ {\left(\sqrt{15 - 2x}\right)}^{2}\hfill&={\left(x\right)}^{2}\hfill \\ 15 - 2x\hfill&={x}^{2}\hfill \end{array}
We see that the remaining equation is a quadratic. Set it equal to zero and solve.
0=x2+2x−15=(x+5)(x−3)x=−5x=3\begin{array}{l}0\hfill&={x}^{2}+2x - 15\hfill \\ \hfill&=\left(x+5\right)\left(x - 3\right)\hfill \\ x\hfill&=-5\hfill \\ x\hfill&=3\hfill \end{array}
The proposed solutions are x=−5x=-5 and x=3x=3. Let us check each solution back in the original equation. First, check x=−5x=-5.
15−2x=x15−2(−5)=−525=−55≠−5\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(-5\right)}\hfill&=-5\hfill \\ \sqrt{25}\hfill&=-5\hfill \\ 5\hfill&\ne -5\hfill \end{array}
This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation. Check x=3x=3.
15−2x=x15−2(3)=39=33=3\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(3\right)}\hfill&=3\hfill \\ \sqrt{9}\hfill&=3\hfill \\ 3\hfill&=3\hfill \end{array}
The solution is x=3x=3.

Try It

Solve the radical equation: x+3=3x−1\sqrt{x+3}=3x - 1

Answer: x=1x=1; extraneous solution x=−29x=-\frac{2}{9}

Solve Equations With Rational Exponents

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, 1612{16}^{\frac{1}{2}} is another way of writing 16\sqrt{16}; 813{8}^{\frac{1}{3}} is another way of writing  83\text{ }\sqrt[3]{8}. 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\frac{2}{3}\left(\frac{3}{2}\right)=1, 3(13)=13\left(\frac{1}{3}\right)=1, and so on.

A General Note: 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=amn=(an)m{a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}

Example: Evaluating a Number Raised to a Rational Exponent

Evaluate 823{8}^{\frac{2}{3}}.

Answer: Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite 823{8}^{\frac{2}{3}} as (813)2{\left({8}^{\frac{1}{3}}\right)}^{2}.

(813)2=(2)2=4\begin{array}{l}{\left({8}^{\frac{1}{3}}\right)}^{2}\hfill&={\left(2\right)}^{2}\hfill \\ \hfill&=4\hfill \end{array}

Try It

Evaluate 64−13{64}^{-\frac{1}{3}}.

Answer: 14\frac{1}{4}

Example: Solving an Equation Involving Rational Exponents and Factoring

Solve 3x34=x123{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.

Answer: This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.

3x34−(x12)=x12−(x12)3x34−x12=0\begin{array}{l}3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)\hfill&={x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}\hfill&=0\hfill \end{array}
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite x12{x}^{\frac{1}{2}} as x24{x}^{\frac{2}{4}}. Then, factor out x24{x}^{\frac{2}{4}} from both terms on the left.
3x34−x24=0x24(3x14−1)=0\begin{array}{l}3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}\hfill&=0\hfill \\ {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill \end{array}
Where did x14{x}^{\frac{1}{4}} come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply x24{x}^{\frac{2}{4}} back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to 24\frac{2}{4} equals 34\frac{3}{4}. Thus, the exponent on x in the parentheses is 14\frac{1}{4}. Let us continue. Now we have two factors and can use the zero factor theorem.
x24(3x14−1)=0x24=0x=03x14−1=03x14=1x14=13Divide both sides by 3.(x14)4=(13)4Raise both sides to the reciprocal of 14.x=181\begin{array}{ll}{x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill & \hfill \\ {x}^{\frac{2}{4}}\hfill&=0\hfill & \hfill \\ x=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}-1\hfill&=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}\hfill&=1\hfill & \hfill \\ {x}^{\frac{1}{4}}\hfill&=\frac{1}{3}\hfill & \text{Divide both sides by 3}.\hfill \\ {\left({x}^{\frac{1}{4}}\right)}^{4}\hfill&={\left(\frac{1}{3}\right)}^{4}\hfill & \text{Raise both sides to the reciprocal of }\frac{1}{4}.\hfill \\ x\hfill&=\frac{1}{81}\hfill & \hfill \end{array}
The two solutions are x=0x=0, x=181x=\frac{1}{81}.

Try It

Solve: (x+5)32=8{\left(x+5\right)}^{\frac{3}{2}}=8.

Answer: {−1}\{-1\}

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 x4−5x2+4=0,x6+7x3−8=0{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0, and x23+4x13+2=0{x}^{\frac{2}{3}}+4{x}^{\frac{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.

A General Note: 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.

  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: Solving a Fourth-Degree Equation in Quadratic Form

Solve this fourth-degree equation: 3x4−2x2−1=03{x}^{4}-2{x}^{2}-1=0.

Answer: This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u=x2u={x}^{2}. Rewrite the equation in u.

3u2−2u−1=03{u}^{2}-2u - 1=0
Now solve the quadratic.
3u2−2u−1=0(3u+1)(u−1)=0\begin{array}{l}3{u}^{2}-2u - 1\hfill=0\hfill \\ \left(3u+1\right)\left(u - 1\right)\hfill=0\hfill \end{array}
Solve each factor and replace the original term for u.
3u+1=03u=−1u=−13x2=−13x=±i13\begin{array}{l}3u+1\hfill&=0\hfill \\ 3u\hfill&=-1\hfill \\ u\hfill&=-\frac{1}{3}\hfill \\ {x}^{2}\hfill&=-\frac{1}{3}\hfill \\ x\hfill&=\pm i\sqrt{\frac{1}{3}}\hfill \end{array}
u−1=0u=1x2=1x=±1\begin{array}{l}u - 1\hfill&=0\hfill \\ u\hfill&=1\hfill \\ {x}^{2}\hfill&=1\hfill \\ x\hfill&=\pm 1\hfill \end{array}
The solutions are x=±i13x=\pm i\sqrt{\frac{1}{3}} and x=±1x=\pm 1.

Try It

Solve using substitution: x4−8x2−9=0{x}^{4}-8{x}^{2}-9=0.

Answer: x=−3,3,−i,ix=-3,3,-i,i

Example: Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form: (x+2)2+11(x+2)−12=0{\left(x+2\right)}^{2}+11\left(x+2\right)-12=0.

Answer: This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u=x+2u=x+2. Then rewrite the equation in u.

u2+11u−12=0(u+12)(u−1)=0\begin{array}{l}{u}^{2}+11u - 12\hfill&=0\hfill \\ \left(u+12\right)\left(u - 1\right)\hfill&=0\hfill \end{array}
Solve using the zero-factor property and then replace u with the original expression.
u+12=0u=−12x+2=−12x=−14\begin{array}{l}u+12\hfill&=0\hfill \\ u\hfill&=-12\hfill \\ x+2\hfill&=-12\hfill \\ x\hfill&=-14\hfill \end{array}
The second factor results in
u−1=0u=1x+2=1x=−1\begin{array}{l}u - 1\hfill&=0\hfill \\ u\hfill&=1\hfill \\ x+2\hfill&=1\hfill \\ x\hfill&=-1\hfill \end{array}
We have two solutions: x=−14x=-14, x=−1x=-1.

Try It

Solve: (x−5)2−4(x−5)−21=0{\left(x - 5\right)}^{2}-4\left(x - 5\right)-21=0.

Answer: x=2,x=12x=2,x=12

Key Concepts

  • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1.
  • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping.
  • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index.
  • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value.
  • Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve.
  • Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form.

Glossary

absolute value equation an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression equations in quadratic form equations with a power other than 2 but with a middle term with an exponent that is one-half the exponent of the leading term extraneous solutions any solutions obtained that are not valid in the original equation polynomial equation an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents radical equation an equation containing at least one radical term where the variable is part of the radicand

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