### Learning Objectives

In this section you will:

We have solved linear equations, exponential 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 rational equations and radical equations. 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.

### 6.3.1 – Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear or quadratic equation. If an equation contains at least one rational expression, it is a considered a rational equation.

Recall that a rational number is the ratio of two numbers, such as $\,\frac{2}{3}\,$ or $\,\frac{7}{2}.\,$ A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

$\frac{x+1}{{x}^{2}-4},\,\frac{1}{x-3},\,\text{or}\,\frac{4}{{x}^{2}+x-2}$

Rational equations have a variable in the denominator in at least one of the terms.
Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

### Example 1 – Solving a Rational Equation

Solve the rational equation: $\,\frac{7}{2x}-\frac{5}{3x}=\frac{22}{3}.$

We have three denominators; $\,2x,3x,$ and 3. The LCD must contain $\,2x,3x,$ and 3. An LCD of $\,6x\,$ contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD $\,6x.$

$\begin{array}{cccc}\hfill \left(6x\right)\left[\frac{7}{2x}-\frac{5}{3x}\right]& =& \left[\frac{22}{3}\right]\left(6x\right)\hfill & \\ \hfill \left(6x\right)\left(\frac{7}{2x}\right)-\left(6x\right)\left(\frac{5}{3x}\right)& =& \left(\frac{22}{3}\right)\left(6x\right)\hfill & \phantom{\rule{2em}{0ex}}\text{Use the distributive property}.\hfill \\ \hfill 3\left(7\right)-2\left(5\right)& =& 22\left(2x\right)\hfill& \phantom{\rule{2em}{0ex}}\text{Cancel out the common factors}.\hfill \\\hfill 21-10& =& 44x\hfill & \\ \hfill 11& =& 44x\hfill & \\ \hfill \frac{11}{44}& =& x\hfill & \\ \hfill \frac{1}{4}& =& x\hfill & \end{array}$

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as $\,\left(x+1\right).\,$ Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are $\,x,$ $x-1,$ and $\,3x-3.\,$ First, factor all denominators. We then have $\,x,$ $\left(x-1\right),$ and $\,3\left(x-1\right)\,$ as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of $\,\left(x-1\right).\,$ The $\,x\,$ in the first denominator is separate from the $\,x\,$ in the $\,\left(x-1\right)\,$ denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the $\,x,$ one factor of $\,\left(x-1\right),$ and the 3. Thus, the LCD is the following:

$x\left(x-1\right)3=3x\left(x-1\right)$

So, both sides of the equation would be multiplied by $\,3x\left(x-1\right).\,$ Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as $\,x\,$ and $\,{x}^{2}+2x.\,$ Once the second denominator is factored as $\,{x}^{2}+2x=x\left(x+2\right),$ there is a common factor of x in both denominators and the LCD is $\,x\left(x+2\right).$

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

$\frac{a}{b}=\frac{c}{d}$

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply $\,a\left(d\right)\,$ and $\,b\left(c\right),$ which results in $\,ad=bc.$

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

### Rational Equations

A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.

### How To

Given a rational equation, solve it.

1. Factor all denominators in the equation.
2. Find and exclude values that set each denominator equal to zero.
3. Find the LCD.
4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
5. Solve the remaining equation.
6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator

### Example 2 – Solving a Rational Equation without Factoring

Solve the following rational equation:

$\frac{2}{x}-\frac{3}{2}=\frac{7}{2x}$

We have three denominators: $\,x,$ $2,$ and $\,2x.\,$ No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is $\,2x.\,$ Only one value is excluded from a solution set, 0.  Next, multiply the whole equation (both sides of the equal sign) by $\,2x.$

$\begin{array}{cccc}\hfill 2x\left[\frac{2}{x}-\frac{3}{2}\right]& =& \left[\frac{7}{2x}\right]2x& \\ \hfill 2x\left(\frac{2}{x}\right)-2x\left(\frac{3}{2}\right)& =& \left(\frac{7}{2x}\right)2x\hfill & \phantom{\rule{2em}{0ex}}\text{Distribute }2x\text{.}\hfill \\ \hfill 2\left(2\right)-3x& =& 7\hfill & \phantom{\rule{2em}{0ex}}\text{Denominators cancel out}.\hfill \\ \hfill 4-3x& =& 7\hfill & \\ \hfill -3x& =& 3\hfill & \\ \hfill x& =& -1\hfill & \\ & & \text{or}\phantom{\rule{0.3em}{0ex}}\left\{-1\right\}\hfill & \end{array}$

The proposed solution is −1,  which is not an excluded value, so the solution set contains one number, $\,x=-1,$ or $\,\left\{-1\right\}\,$ written in set notation.

### Try It

Solve the rational equation: $\,\frac{2}{3x}=\frac{1}{4}-\frac{1}{6x}.$

$x=\frac{10}{3}$

### Example 3 – Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: $\,\frac{1}{x}=\frac{1}{10}-\frac{3}{4x}.$

First find the common denominator. The three denominators in factored form are $\,x,10=2\cdot 5,$ and $\,4x=2\cdot 2\cdot x.\,$ The smallest expression that is divisible by each one of the denominators is $\,20x.\,$ Only $\,x=0\,$ is an excluded value. Multiply the whole equation by $\,20x.$

$\begin{array}{ccc}\hfill 20x\left(\frac{1}{x}\right)& =& \left(\frac{1}{10}-\frac{3}{4x}\right)20x\hfill \\ \hfill 20& =& 2x-15\hfill \\ \hfill 35& =& 2x\hfill \\ \hfill \frac{35}{2}& =& x\hfill \end{array}$

The solution is $\,\frac{35}{2}.$

### Try It

Solve the rational equation: $\,-\frac{5}{2x}+\frac{3}{4x}=-\frac{7}{4}.$

$x=1$

### Example 4 – Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

1. $\frac{3}{x-6}=\frac{5}{x}$
2. $\frac{x}{x-3}=\frac{5}{x-3}-\frac{1}{2}$
3. $\frac{x}{x-2}=\frac{5}{x-2}-\frac{1}{2}$
1. The denominators $\,x\,$ and $\,x-6\,$ have nothing in common. Therefore, the LCD is the product $\,x\left(x-6\right).\,$ However, for this problem, we can cross-multiply.

$\begin{array}{cccc}\hfill \frac{3}{x-6}& =& \frac{5}{x}\hfill & \\ \hfill 3x& =& 5\left(x-6\right)\hfill & \phantom{\rule{2em}{0ex}}\text{Distribute}\text{.}\hfill \\ \hfill 3x& =& 5x-30\hfill & \\ \hfill -2x& =& -30\hfill & \\ \hfill x& =& 15\hfill & \end{array}$

The solution is 15.  The excluded values are $6$ and $0.$

2. The LCD is $\,2\left(x-3\right).\,$ Multiply both sides of the equation by $\,2\left(x-3\right).$

$\begin{array}{ccc}\hfill 2\left(x-3\right)\left[\frac{x}{x-3}\right]& =& \left[\frac{5}{x-3}-\frac{1}{2}\right]2\left(x-3\right)\hfill \\ \hfill \frac{2\left(x-3\right)x}{x-3}& =& \frac{2\left(x-3\right)5}{x-3}-\frac{2\left(x-3\right)}{2}\hfill \\ \hfill 2x& =& 10-\left(x-3\right)\hfill \\ \hfill 2x& =& 10-x+3\hfill \\ \hfill 2x& =& 13-x\hfill \\ \hfill 3x& =& 13\hfill \\ \hfill x& =& \frac{13}{3}\hfill \end{array}$

The solution is $\,\frac{13}{3}.\,$ The excluded value is $3.$

3. The least common denominator is $\,2\left(x-2\right).\,$ Multiply both sides of the equation by $\,x\left(x-2\right).$

$\begin{array}{ccc}\hfill 2\left(x-2\right)\left[\frac{x}{x-2}\right]& =& \left[\frac{5}{x-2}-\frac{1}{2}\right]2\left(x-2\right)\hfill \\ \hfill 2x& =& 10-\left(x-2\right)\hfill \\ \hfill 2x& =& 12-x\hfill \\ \hfill 3x& =& 12\hfill \\ \hfill x& =& 4\hfill \end{array}$

The solution is 4. The excluded value is $2.$

### Try It

Solve $\,\frac{-3}{2x+1}=\frac{4}{3x+1}.\,$ State the excluded values.

$x=-\frac{7}{17}.\,$ Excluded values are $\,x=-\frac{1}{2}\,$ and $\,x=-\frac{1}{3}.$

### Example 5 – Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: $\,\frac{2}{x+1}-\frac{1}{x-1}=\frac{2x}{{x}^{2}-1}.\,$ State the excluded values.

We must factor the denominator $\,{x}^{2}-1.\,$ We recognize this as the difference of squares, and factor it as $\,\left(x-1\right)\left(x+1\right).\,$ Thus, the LCD that contains each denominator is $\,\left(x-1\right)\left(x+1\right).\,$ Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.

$\begin{array}{ccc}\hfill \left(x-1\right)\left(x+1\right)\left[\frac{2}{x+1}-\frac{1}{x-1}\right]& =& \left[\frac{2x}{\left(x-1\right)\left(x+1\right)}\right]\left(x-1\right)\left(x+1\right)\hfill \\ \hfill 2\left(x-1\right)-1\left(x+1\right)& =& 2x\hfill \\ \hfill 2x-2-x-1& =& 2x\phantom{\rule{2em}{0ex}}\text{Distribute the negative sign}.\hfill \\ \hfill -3-x& =& 0\hfill \\ \hfill -3& =& x\hfill \end{array}$

The solution is $\,-3.\,$ The excluded values are $\,1\,$ and $\,-1.$

### Try It

Solve the rational equation: $\,\frac{2}{x-2}+\frac{1}{x+1}=\frac{1}{{x}^{2}-x-2}.$

$x=\frac{1}{3}$

### Example 6 – Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: $\,\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.$

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, $\,{x}^{2}-1=\left(x+1\right)\left(x-1\right).\,$ Then, the LCD is $\,\left(x+1\right)\left(x-1\right).\,$ Next, we multiply the whole equation by the LCD.

$\begin{array}{ccc}\hfill \left(x+1\right)\left(x-1\right)\left[\frac{-4x}{x-1}+\frac{4}{x+1}\right]& =& \left[\frac{-8}{\left(x+1\right)\left(x-1\right)}\right]\left(x+1\right)\left(x-1\right)\hfill \\ \hfill -4x\left(x+1\right)+4\left(x-1\right)& =& -8\hfill \\ \hfill -4{x}^{2}-4x+4x-4& =& -8\hfill \\ \hfill -4{x}^{2}+4& =& 0\hfill \\ \hfill -4\left({x}^{2}-1\right)& =& 0\hfill \\ \hfill -4\left(x+1\right)\left(x-1\right)& =& 0\hfill \\ \hfill x& =& -1\hfill \\ \hfill x& =& 1\hfill \end{array}$

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

### Try It

Solve $\,\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.$

$x=-1,$ $x=0$ is not a solution.

### 6.3.2 – Solving Radical Equations

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

$\begin{array}{ccc}\hfill \sqrt{3x+18}& =& x\hfill \\ \hfill \sqrt{x+3}& =& x-3\hfill \\ \hfill \sqrt{x+5}-\sqrt{x-3}& =& 2\hfill \end{array}$

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.

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 7 – Solving an Equation with One Radical

Solve $\,\sqrt{15-2x}=x.$

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

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill {\left(\sqrt{15-2x}\right)}^{2}& =& {\left(x\right)}^{2}\hfill \\ \hfill 15-2x& =& {x}^{2}\hfill \end{array}$

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

$\begin{array}{ccc}\hfill 0& =& {x}^{2}+2x-15\hfill \\ & =& \left(x+5\right)\left(x-3\right)\hfill \\ \hfill x& =& -5\hfill \\ \hfill x& =& 3\hfill \end{array}$

The proposed solutions are $-5\,$ and $3.\,$ Let us check each solution back in the original equation. First, check $\,x=-5.$

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill \sqrt{15-2\left(-5\right)}& =& -5\hfill \\ \hfill \sqrt{25}& =& -5\hfill \\ \hfill 5& \ne & -5\hfill \end{array}$

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 $\,x=3.$

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill \sqrt{15-2\left(3\right)}& =& 3\hfill \\ \hfill \sqrt{9}& =& 3\hfill \\ \hfill 3& =& 3\hfill \end{array}$

The solution is $\,3.$

### Try It

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

$x=1;$ extraneous solution $\,x=-\frac{2}{9}$

Solve $\,\sqrt{2x+3}+\sqrt{x-2}=4.$

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

$\begin{array}{cccc}\hfill \sqrt{2x+3}+\sqrt{x-2}& =& 4\hfill & \\ \hfill \sqrt{2x+3}& =& 4-\sqrt{x-2}\hfill & \phantom{\rule{2em}{0ex}}\text{Subtract }\sqrt{x-2}\text{ from both sides}.\hfill \\ \hfill {\left(\sqrt{2x+3}\right)}^{2}& =& {\left(4-\sqrt{x-2}\right)}^{2}\hfill & \phantom{\rule{2em}{0ex}}\text{Square both sides}.\hfill \end{array}$

Use the perfect square formula to expand the right side: $\,{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}.$

$\begin{array}{cccc}\hfill 2x+3& =& {\left(4\right)}^{2}-2\left(4\right)\sqrt{x-2}+{\left(\sqrt{x-2}\right)}^{2}\hfill & \\ \hfill 2x+3& =& 16-8\sqrt{x-2}+\left(x-2\right)\hfill & \\ \hfill 2x+3& =& 14+x-8\sqrt{x-2}\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ \hfill x-11& =& -8\sqrt{x-2}\hfill & \phantom{\rule{2em}{0ex}}\text{Isolate the second radical}.\hfill \\ \hfill {\left(x-11\right)}^{2}& =& {\left(-8\sqrt{x-2}\right)}^{2}\hfill & \phantom{\rule{2em}{0ex}}\text{Square both sides}.\hfill \\ \hfill {x}^{2}-22x+121& =& 64\left(x-2\right)\hfill & \end{array}$

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

$\begin{array}{cccc}\hfill {x}^{2}-22x+121& =& 64x-128\hfill & \\ \hfill {x}^{2}-86x+249& =& 0\hfill & \\ \hfill \left(x-3\right)\left(x-83\right)& =& 0\hfill & \phantom{\rule{2em}{0ex}}\text{Factor and solve}.\hfill \\ \hfill x& =& 3\hfill & \\ \hfill x& =& 83\hfill & \end{array}$

The proposed solutions are $\,3\,$ and $\,83.\,$ Check each solution in the original equation.

$\begin{array}{ccc}\hfill \sqrt{2x+3}+\sqrt{x-2}& =& 4\hfill \\ \hfill \sqrt{2x+3}& =& 4-\sqrt{x-2}\hfill \\ \hfill \sqrt{2\left(3\right)+3}& =& 4-\sqrt{\left(3\right)-2}\hfill \\ \hfill \sqrt{9}& =& 4-\sqrt{1}\hfill \\ \hfill 3& =& 3\hfill \end{array}$

One solution is $\,3.$

Check $\,x=83.$

$\begin{array}{ccc}\hfill \sqrt{2x+3}+\sqrt{x-2}& =& 4\hfill \\ \hfill \sqrt{2x+3}& =& 4-\sqrt{x-2}\hfill \\ \hfill \sqrt{2\left(83\right)+3}& =& 4-\sqrt{\left(83-2\right)}\hfill \\ \hfill \sqrt{169}& =& 4-\sqrt{81}\hfill \\ \hfill 13& \ne & -5\hfill \end{array}$

The only solution is $\,3.\,$ We see that $\,x=83\,$ is an extraneous solution.

### Try It

Solve the equation with two radicals: $\,\sqrt{3x+7}+\sqrt{x+2}=1.$

$x=-2;$ extraneous solution $\,x=-1$

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

### Key Concepts

• A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See Example 1 and Example 2.
• All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See Example 3, Example 4, and Example 5.
• Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example 6.
• We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See Example 7 and Example 8.

### Glossary

rational equation
an equation consisting of a fraction of polynomials
extraneous solutions
any solutions obtained that are not valid in the original equation