Lesson 3: Solving Radical Equations

Property

An equation in which the variable is in the radicand of a square root is called a radical equation.
When we solve radical equations by squaring both sides we may get an algebraic solution that would make $\sqrt{a}$ negative.
This algebraic solution would not be a solution to the original radical equation; it is an extraneous solution.

Examples

• For the equation $\sqrt{x+6} = x$, is $x=3$ a solution? Yes, because $\sqrt{3+6} = \sqrt{9} = 3$.

• For the same equation, $\sqrt{x+6} = x$, is $x=-2$ a solution? No, because $\sqrt{-2+6} = \sqrt{4} = 2$, and $2 \ne -2$. So, $x=-2$ is an extraneous solution.

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 $n=13$.