Solving Rational Equations

Rational equations can be solved by algebraic methods or by graphing.

Solving a radical equation by graphing is the same as solving a rational equation by graphing. For equations of the form $f(x)=0$, graph $f(x)$ and find the zeros (the points where the graph intersects the $x$-axis). For equations of the form $f(x)=g(x)$, graph $f(x)$ and $g(x)$ and find the $x$-values of the points where the graphs intersect. These $x$-values are the solutions of the equation. If technology is used to graph the functions, then zero, trace, or intersect features may be used to find the solutions.

To solve rational equations algebraically, write the equation in the form:
$\frac{p(x)}{q(x)}=0$
Then $x$ is a solution if $p(x)=0$ and $q(x)\neq0$.
Step-By-Step Example
Solving Rational Equations by Graphing
Graph the equation:
$\frac{x^2-9}{x^2-4x+3}=0$
Step 1

Identify the vertical asymptote and horizontal asymptote.

The graph has a vertical asymptote at $x=1$, a hole at $x=3$, and a horizontal asymptote at $y=1$.
Step 2

Identify the zeros of the graph. The zeros of the numerator are $x=3$ and $x=-3$. However, the denominator is also zero at $x=3$, so that is not a valid replacement value for the variable.

The graph shows only one zero, at $x=-3$.

Solution
The solution is $x=-3$.
Step-By-Step Example
Solving Rational Equations Algebraically
Solve the equation:
$\frac{x+1}{x+2}=\frac{2}{x^2+2x}$
Step 1
Write the equation in the form:
$\frac{p(x)}{q(x)}=0$
Set the equation equal to zero by subtracting one expression from both sides:
$\frac{x+1}{x+2}-\frac{2}{x^2+2x}=0$
Identify a common denominator.
$\frac{x+1}{x+2}\left(\frac{x}{x}\right)-\frac{2}{x^2+2x}=0$
Simplify.
$\frac{x^2+x}{x^2+2x}-\frac{2}{x^2+2x}=0$
Combine into one fraction.
$\frac{x^2+x-2}{x^2+2x}=0$
Step 2
Identify the zeros of the numerator. Start by factoring.
\begin{aligned}x^2+x-2&=0\\(x+2)(x-1)&=0\end{aligned}
Use the zero product property (if $pq=0$, then $p=0$ or $q=0$) for each factor and solve.
\begin{aligned} x+2&=0\\x&=-2\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned} x-1&=0\\x&=1\end{aligned}
Step 3

Check whether the zeros of the numerator are also zeros of the denominator in the equation:

Substitute –2 for $x$.
$\begin{gathered}x^2+2x\\(-2)^2+2(-2)\\4-4\\0\end{gathered}$
So, –2 is not a valid replacement value of the variable. Substitute 1 for $x$.
$\begin{gathered}x^{2}+2x\\1^2+2(1)\\1+2\\3\end{gathered}$
The value $x=1$ is a zero of the numerator, but not of the denominator. So, it is a solution.
Solution
The solution is $x=1$.

Radical equations can be solved by algebraic methods or by graphing.

Solving a radical equation by graphing is the same as solving a rational equation by graphing.

• For equations of the form $f(x)=0$, graph $f(x)$ and find the zeros.
• For equations of the form $f(x)=g(x)$, graph $f(x)$ and $g(x)$ and find the $x$-values of the points where the graphs intersect.
Step-By-Step Example
$\sqrt{x^3-4x+5}=2$
Step 1
Graph two related functions:
\begin{aligned}f(x)&=\sqrt{x^3-4x+5}\\\\g(x)&=2\end{aligned}
Step 2

Identify the points where the graphs intersect. When using technology, the values may be estimates.

The graphs intersect at approximately $(-2.11,2)$, $(0.25,2)$, and $(1.86,2)$.

Solution
The solutions of the radical equation are: $x\approx-2.11$, $x\approx0.25$, and $x\approx1.86$.

An extraneous solution is a solution of a step in solving that is not a solution of the original equation. When solving radical equations by squaring both sides, extraneous solutions can be introduced. For example, if both sides of $\sqrt{x}=-3$ are squared, the result is $x=9$, which is not a solution of the original equation because $\sqrt{9}\neq-3$. Therefore, all solutions must be checked in the original equation.

To solve radical equations algebraically, the method depends on the structure of the equation.

For equations with one radical, $\sqrt{f(x)}=g(x)$:

1. Isolate the radical on one side of the equation.

2. Square both sides to eliminate the radical.

3. Solve the resulting equation.

4. Check for extraneous solutions.

For equations with two radicals and no terms outside the radical, $\sqrt{f(x)}=\sqrt{g(x)}$:

1. Write the equation with one radical on each side.

2. Square both sides to eliminate the radicals.

3. Solve the resulting equation.

4. Check for extraneous solutions.

For equations with two radicals and terms outside the radicals, $\sqrt{f(x)}=\sqrt{g(x)}+h(x)$:

1. Write the equation with one radical isolated on one side.

2. Square both sides of the equation.

3. Isolate the remaining radical on one side of the equation.

4. Square both sides to eliminate the radical.

5. Solve the resulting equation.

6. Check for extraneous solutions.

To solve equations with other roots, raise both sides to the power that is the inverse of the root.

Step-By-Step Example
Solve the equation:
$\sqrt{x^2-x-6}=6$
Step 1
$\sqrt{x^2-x-6}=6$
Step 2
Square both sides to eliminate the radical.
\begin{aligned}(\sqrt{x^2-x-6})^2&=6^2\\x^2-x-6&=36\end{aligned}
Step 3
The standard form of an equation is:
$ax^{2}+bx+c=0$
Write the resulting equation in standard form by subtracting 36 on both sides of the equation:
\begin{aligned}x^{2}-x-6&=36\\ x^{2}-x-6-36&=36-36\\x^{2}-x-42&=0\end{aligned}
Step 4
Factor the equation.
\begin{aligned}x^{2}-x-42&=0 \\ (x-7)(x+6)&=0\end{aligned}
Step 5
Apply the zero product property by setting each factor equal to zero.
$(x-7)=0\;\;\;\;\;\text{or}\;\;\;\;\;(x+6)=0$
Solve each equation.
\begin{aligned}x-7&=0\\x-7+7&=0+7\\x&=7\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x+6&=0\\x+6-6&=0-6\\x&=-6\end{aligned}
Solution

Check for extraneous solutions.

Substitute $x=7$:
\begin{aligned}\sqrt{(7)^{2}-7-6}&=6\\\sqrt{36}&=6\;\;\checkmark\end{aligned}
Substitute $x=-6$:
\begin{aligned}\sqrt{(-6)^2-(-6)-6}&=6\\\sqrt{36}&=6\;\;\checkmark\end{aligned}
There are no extraneous solutions. Therefore, both $x=7$ and $x=-6$ are valid solutions.

Note: Do not assume that a solution is extraneous because it is negative.

Step-By-Step Example
Solve the equation:
$\sqrt{x+14}=\sqrt{5-2x}$
Step 1
Square both sides of the equation, and solve for $x$.
\begin{aligned}\left(\sqrt{x+14}\right)^2&=\left(\sqrt{5-2x}\right)^2\\x+14&=5-2x\\3x&=-9\\x&=-3\end{aligned}
Step 2

Check the solution in the original equation to make sure it is not extraneous.

Substitute $x=-3$:
\begin{aligned}\sqrt{(-3)+14}&\stackrel{?}{=}\sqrt{5-2(-3)}\\\sqrt{11}&\stackrel{?}{=}\sqrt{5+6}\\\sqrt{11}&=\sqrt{11}\;\;\checkmark\end{aligned}
Solution
$x=-3$ is a valid solution.
Step-By-Step Example
Solving Equations with Two Radicals and Other Terms
Solve the equation:
$\sqrt{3-3x}-\sqrt{x}=1$
Step 1
Write the equation with one radical isolated on one side.
$\sqrt{3-3x}=\sqrt{x}+1$
Step 2
Square both sides of the equation.
\begin{aligned}\left(\sqrt{3-3x}\right)^2&=\left(\sqrt{x}+1\right)^2\\3-3x&=\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\\3-3x&=\left(\sqrt{x}\right)^2+\sqrt{x}+\sqrt{x}+1\\3-3x&=x+2\sqrt{x}+1\end{aligned}
Step 3
Isolate the remaining radical on one side of the equation.
\begin{aligned}2-4x&=2\sqrt{x}\\1-2x&=\sqrt{x}\end{aligned}
Step 4
Square both sides to eliminate the radical.
\begin{aligned}\left(1-2x\right)^2&=\left(\sqrt{x}\right)^2\\1-4x+4x^2&=x\end{aligned}
Step 5
The standard form of an equation is:
$ax^{2}+bx+c=0$
Write the resulting equation in standard form by subtracting $x$ from both sides of the equation:
\begin{aligned}1-4x+4x^{2}&=x\\1-4x+4x^{2}-x&=x-x\\1-5x+4x^{2}&=0\\4x^{2}-5x+1&=0\end{aligned}
Step 6
Factor the equation.
\begin{aligned}4x^2-5x+1&=0\\(4x-1)(x-1)&=0\end{aligned}
Step 7
Apply the zero product property by setting each factor equal to zero.
$(4x-1)=0\;\;\;\;\; \text{or} \;\;\;\;\;(x-1)=0$
Solve each equation.
\begin{aligned}4x-1&=0\\4x-1+1&=0+1\\4x&=1\\\frac{4x}{4}&=\frac{1}{4}\\x&=\frac{1}{4}\end{aligned}\;\;\;\;\; \text{or} \;\;\;\;\;\begin{aligned}x-1&=0\\x-1+1&=0+1\\x&=1\end{aligned}
Solution

Check for extraneous solutions.

Substitute $x=\frac{1}{4}$:
\begin{aligned}\sqrt{3-3\left(\frac{1}{4}\right)}-\sqrt{\frac{1}{4}}&\stackrel{?}{=}1\\\sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}&\stackrel{?}{=}1\\\frac{3}{2}-\frac{1}{2}&=1\;\;\checkmark\end{aligned}
Substitute $x=1$:
\begin{aligned}\sqrt{3-3(1)}-\sqrt{1}&\stackrel{?}{=}1\\\sqrt{0}-\sqrt{1}&\stackrel{?}{=}1\\0-1&\neq1\end{aligned}
$x=1$ is an extraneous solution. So, $x=\frac{1}{4}$ is the only valid solution for the equation.