Raising Equations to Powers: Practice

Work inside the root first, then take the principal (nonnegative) square root.

$\sqrt{5^2} = \sqrt{25} = 5$

The radical returns only the nonnegative root, so the answer is $5$.

Square the $-8$ first, which makes it positive, then take the principal root.

$\sqrt{(-8)^2} = \sqrt{64} = 8$

The answer is $8 = \lvert -8 \rvert$, not $-8$: squaring forgets the sign, and the root gives back only the nonnegative value.

Square both sides to undo the square root.

$x = 9^2 = 81$

Check: $\sqrt{81} = 9$, so $x = 81$.

Square both sides, then solve the linear equation.

$x - 1 = 16 \quad\Longrightarrow\quad x = 17$

Check: $\sqrt{17 - 1} = \sqrt{16} = 4$.

Equal quantities stay equal when you do the same thing to both, so squaring both sides is valid.

$a = b \quad\Longrightarrow\quad a^2 = b^2$

The reverse direction need not hold, but this forward direction always does.

Squaring forgets the sign, so the squared equation gains an extra root.

$x^2 = 25 \quad\Longrightarrow\quad x = 5 \ \text{ or } \ x = -5$

The equation $x = 5$ has just one solution, so squaring added $x = -5$.

Square both sides, then subtract $5$.

$x + 5 = 9 \quad\Longrightarrow\quad x = 4$

Check: $\sqrt{4 + 5} = \sqrt{9} = 3$.

Squaring is not reversible, so the squared equation can gain roots the original never had. For example, $x = 2$ has one solution, but squaring gives

$x^2 = 4 \quad\Longrightarrow\quad x = 2 \ \text{ or } \ x = -2,$

which adds $x = -2$. Substituting each candidate into the original equation removes any such extras.

Taking an even root of both sides brings back both signs.

$x^2 = 49 \quad\Longrightarrow\quad x = \pm 7$

Both $7$ and $-7$ square to $49$, so both are solutions.

Square both sides to clear the root.

$x = 0^2 = 0$

Check: $\sqrt{0} = 0$, so $x = 0$ is the only solution.

Squaring undoes the square root exactly, because $\sqrt{7}$ is the number that squares to $7$.

$\left(\sqrt{7}\right)^2 = 7$

This is why squaring both sides is the move that clears a square root.

An odd power keeps $a^n = b^n$ equivalent to $a = b$, so it cannot create a new branch.

$a^3 = b^3 \quad\Longleftrightarrow\quad a = b$

Squaring and the fourth and sixth powers are even, and an even power can add solutions.