Lesson 4: Solving Radical Equations and Inequalities

Property

To solve a radical equation:

1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve the new equation.
4. Check the answer.

This strategy is based on the property that for $a \ge 0$, $(\sqrt{a})^2 = a$.

Examples

• To solve $\sqrt{3x-2} = 5$, square both sides: $(\sqrt{3x-2})^2 = 5^2$, which gives $3x-2 = 25$. Solving for $x$ gives $3x = 27$, so $x=9$.

• To solve $\sqrt{4n-3} - 7 = 0$, first isolate the radical: $\sqrt{4n-3} = 7$. Square both sides: $4n-3 = 49$. This gives $4n=52$, so $n=13$.

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.