Math Core

A radical equation has the variable inside a radical, like $\sqrt{x + 7} = x + 1$ . The strategy is simple: isolate the radical and raise both sides to a power to remove it. The catch is that this step can create answers that don't actually work, so checking is not optional here.

The basic method

Since $\left(\sqrt[n]{u}\right)^n = u$ , raising a radical to its index removes the radical. That gives a four-step method.

Solving a radical equation

Isolate the radical on one side.
Raise both sides to the power of the index (square for a square root, cube for a cube root).
Solve the resulting equation.
Check every solution in the original equation. Reject any that fail.

Worked example: One radical, one solution

Solve $\sqrt{2x + 3} - 1 = 4$ .

Isolate the radical: $\sqrt{2x + 3} = 5$ . Square both sides:

\begin{aligned} 2x + 3 &= 25 \\ 2x &= 22 \\ x &= 11 \end{aligned}

Check: $\sqrt{2(11) + 3} - 1 = \sqrt{25} - 1 = 5 - 1 = 4$ . ✓ The solution is $x = 11$ .

Always isolate the radical before squaring. If you square $\sqrt{2x + 3} - 1 = 4$ directly, the left side becomes $\left(\sqrt{2x + 3} - 1\right)^2 = 2x + 3 - 2\sqrt{2x + 3} + 1$ , and the radical is still there.

Why extraneous solutions appear

Squaring is not a reversible step. If $a = b$ , then $a^2 = b^2$ . But if $a^2 = b^2$ , you only know $a = b$ or $a = -b$ . For example, $3 \ne -3$ , but $3^2 = (-3)^2$ . So squaring both sides can turn a false equation into a true one, and the new equation may have solutions the original doesn't.

Definition

Extraneous solution

An extraneous solution is a value that solves an equation produced during the solving process but does not solve the original equation.

Extraneous solutions usually come from a sign problem: a square root is never negative, so any candidate that makes the other side of $\sqrt{\ \ } = \ldots$ negative must be rejected.

Worked example: An extraneous solution

Solve $\sqrt{x + 7} = x + 1$ .

The radical is already isolated. Square both sides and solve the quadratic:

\begin{aligned} x + 7 &= (x + 1)^2 \\ x + 7 &= x^2 + 2x + 1 \\ 0 &= x^2 + x - 6 \\ 0 &= (x + 3)(x - 2) \end{aligned}

The candidates are $x = -3$ and $x = 2$ . Check both in the original equation:

$x = 2$ : left side $\sqrt{9} = 3$ , right side $2 + 1 = 3$ . ✓
$x = -3$ : left side $\sqrt{4} = 2$ , right side $-3 + 1 = -2$ . ✗

So $x = -3$ is extraneous, and the only solution is $x = 2$ .

The graph shows what happened. The solutions of $\sqrt{x + 7} = x + 1$ are the $x$ -coordinates where the two graphs meet, and they meet only once, at $(2, 3)$ . The extraneous value $x = -3$ solves $\sqrt{x + 7} = -(x + 1)$ instead, the equation that squaring mixed in.

y = √(x + 7) and y = x + 1 intersect only at (2, 3).Open in grapher →

Common mistake

Never skip the check. A candidate can come out of perfectly correct algebra and still be wrong. Also, if an isolated square root equals a negative number, such as $\sqrt{x - 3} = -2$ , you can stop right away: there is no solution, since a principal square root is never negative.

Cube roots and other odd indices

Cubing is reversible: if $a^3 = b^3$ , then $a = b$ . So equations with odd-index radicals don't produce extraneous solutions (it's still smart to check for arithmetic errors).

Worked example: A cube root equation

Solve $\sqrt[3]{x - 4} + 3 = 1$ .

Isolate: $\sqrt[3]{x - 4} = -2$ . A cube root can be negative, so keep going. Cube both sides:

x - 4 = (-2)^3 = -8, \qquad x = -4.

Check: $\sqrt[3]{-4 - 4} + 3 = \sqrt[3]{-8} + 3 = -2 + 3 = 1$ . ✓

Equations with two radicals

When there are two square roots, isolate one of them, square, and then isolate the remaining radical and square again.

Worked example: Two radicals

Solve $\sqrt{x + 5} - \sqrt{x} = 1$ .

Isolate one radical: $\sqrt{x + 5} = 1 + \sqrt{x}$ . Square both sides. The right side is a binomial, so it has a middle term:

\begin{aligned} x + 5 &= 1 + 2\sqrt{x} + x \\ 4 &= 2\sqrt{x} \\ 2 &= \sqrt{x} \\ 4 &= x \end{aligned}

Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ . ✓ The solution is $x = 4$ .

Equations with rational exponents

An equation like $(x + 2)^{3/2} = 8$ is a radical equation in disguise, since $(x + 2)^{3/2} = \left(\sqrt{x + 2}\right)^3$ . Raise both sides to the reciprocal power:

x + 2 = 8^{2/3} = 4, \qquad x = 2.

Check: $(2 + 2)^{3/2} = 4^{3/2} = 8$ . ✓ As in the last lesson, an even numerator means two candidates, and an even denominator means the base can't be negative, so check every value.

An application

The time $T$ (in seconds) for a pendulum of length $L$ (in feet) to swing back and forth once is about $T = 2\pi\sqrt{\dfrac{L}{32}}$ . To find the length that gives a period of $2$ seconds, isolate and square:

\sqrt{\frac{L}{32}} = \frac{2}{2\pi} = \frac{1}{\pi}, \qquad \frac{L}{32} = \frac{1}{\pi^2}, \qquad L = \frac{32}{\pi^2} \approx 3.24 \text{ feet}.

Tip

Before solving, think about the domain. In $\sqrt{x + 7} = x + 1$ , the left side is never negative, so any solution needs $x + 1 \ge 0$ , that is, $x \ge -1$ . That tells you in advance that $x = -3$ can't work.