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.
Idea clave
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).
Por qué funciona
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
Ejemplo resuelto: One root, one step
Solve $\sqrt{2x - 1} = 5$.
- Square: $2x - 1 = 25$, so $x = 13$.
- Check: $\sqrt{25} = 5$.
Respuesta: $x = 13$
Ejemplo resuelto: 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.
Respuesta: $x = 1$
Ejemplo resuelto: 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.
Respuesta: $x = 6$
Error común
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.