Lesson 8.6: Solve Radical Equations

New Concept

A radical equation has a variable in the radicand. The strategy is to isolate the radical and then raise both sides to the power of the index, using the property $(\sqrt[n]{a})^n = a$. Always check for extraneous solutions.

What’s next

You'll work through interactive examples of solving equations with one and two radicals. Then, you will apply these skills in challenge problems and real-world applications.

Property

An equation in which a variable is in the radicand of a radical expression is called a radical equation.

HOW TO: Solve a radical equation with one radical.
Step 1. Isolate the radical on one side of the equation.
Step 2. Raise both sides of the equation to the power of the index.
Step 3. Solve the new equation.
Step 4. Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Property

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. This is called an extraneous solution. Always check your answer in the original equation.

Examples

• Solve $\sqrt{r+1} = r-1$. Squaring gives $r+1 = r^2-2r+1$, which simplifies to $r^2-3r=0$, or $r(r-3)=0$. The algebraic solutions are $r=0$ and $r=3$. Checking $r=0$ gives $\sqrt{1}=-1$ (False). Checking $r=3$ gives $\sqrt{4}=2$ (True). Thus, $r=0$ is an extraneous solution.

• Solve $\sqrt{m+9} - m + 3 = 0$. Isolate the radical: $\sqrt{m+9} = m-3$. Squaring gives $m+9 = m^2-6m+9$, so $m^2-7m=0$. The solutions are $m=0$ and $m=7$. Only $m=7$ is a valid solution.

Property

HOW TO: Solve a radical equation with two radicals.
Step 1. Isolate one of the radical terms on one side of the equation.
Step 2. Raise both sides of the equation to the power of the index.
Step 3. Are there any more radicals? If yes, repeat Step 1 and Step 2 again. If no, solve the new equation.
Step 4. Check the answer in the original equation.

Examples

• Solve $\sqrt[3]{6x-5} = \sqrt[3]{2x+7}$. The radicals are already isolated. Cube both sides: $6x-5 = 2x+7$. This simplifies to $4x=12$, so $x=3$.

• Solve $\sqrt{x-5} + 2 = \sqrt{x+7}$. Square both sides: $(\sqrt{x-5}+2)^2 = x+7$. This gives $(x-5) + 4\sqrt{x-5} + 4 = x+7$. Simplify to $4\sqrt{x-5} = 8$, or $\sqrt{x-5}=2$. Square again: $x-5=4$, so $x=9$.