To solve an equation with a root, undo the root by raising both sides to a power, the inverse step you know from nth Roots and Simplifying and Rationalizing Radicals. Squaring is powerful, but it can create answers that aren't real solutions, so checking is part of the method.
Key idea
Isolate the radical. Raise both sides to the index (square for $\sqrt{\ }$, cube for $\sqrt[3]{\ }$). Solve. Then check every answer in the original equation and throw out any that fail (extraneous solutions).
Why it works
If $a = b$ then $a^2 = b^2$, but the reverse is false: $a^2 = b^2$ also holds when $a = -b$. So squaring adds the solutions of a second equation, the one with the opposite sign. A square root is never negative, so any answer that would make a square root equal a negative number belongs to that other equation and must be rejected. Cubing doesn't have this problem, because $a^3 = b^3$ only when $a = b$.
Worked examples
Worked example: One root, one step
Solve $\sqrt{2x - 1} = 5$.
- Square: $2x - 1 = 25$, so $x = 13$.
- Check: $\sqrt{25} = 5$.
Answer: $x = 13$
Worked example: An extraneous solution
Solve $\sqrt{x + 3} = x + 1$.
- Square: $x + 3 = x^2 + 2x + 1$, so $x^2 + x - 2 = 0$.
- Factor: $(x + 2)(x - 1) = 0$, so $x = -2$ or $x = 1$.
- Check $x = 1$: $\sqrt{4} = 2$ and $1 + 1 = 2$. Works.
- Check $x = -2$: $\sqrt{1} = 1$ but $-2 + 1 = -1$. Fails, so it is extraneous.
Answer: $x = 1$
Worked example: Isolate first
Solve $\sqrt{x - 2} + 4 = x$.
- Isolate: $\sqrt{x - 2} = x - 4$.
- Square: $x - 2 = x^2 - 8x + 16$, so $x^2 - 9x + 18 = 0 = (x - 3)(x - 6)$.
- $x = 6$: $\sqrt{4} = 2 = 6 - 4$. Works. $x = 3$: $\sqrt{1} = 1$ but $3 - 4 = -1$. Extraneous.
Answer: $x = 6$
Common mistake
Squaring term by term. $(x + 1)^2$ is $x^2 + 2x + 1$, not $x^2 + 1$, and you can't square $\sqrt{x - 2} + 4$ by squaring each piece. Isolate the root first, then square whole sides. And always check: skipping the check is how extraneous answers sneak in.