### Solving Rational Equations

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: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 $y=1$.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 $x=-3$.

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

Substitute –2 for $x$.### Solving Radical Equations

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.

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 $(1.86,2)$.

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.

Check for extraneous solutions.

Substitute $x=7$:Note: Do not assume that a solution is extraneous because it is negative.

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

Substitute $x=-3$:Check for extraneous solutions.

Substitute $x=\frac{1}{4}$: