Raising Equations to Powers: Practice

Isolate the radical first by subtracting $2$, then square.

$\sqrt{x} = 5 \quad\Longrightarrow\quad x = 25$

Check: $2 + \sqrt{25} = 2 + 5 = 7$.

Square both sides and collect into a quadratic.

$x + 3 = x^2 + 2x + 1 \quad\Longrightarrow\quad x^2 + x - 2 = 0$

Factoring gives $(x + 2)(x - 1) = 0$, so the candidates are $x = -2$ and $x = 1$. Checking, $x = 1$ gives $\sqrt{4} = 2$ and $1 + 1 = 2$, but $x = -2$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 1$ works.

Split the root and apply $\sqrt{x^2} = \lvert x \rvert$.

$\sqrt{9x^2} = \sqrt{9}\,\sqrt{x^2} = 3\lvert x \rvert$

Writing $3x$ would be wrong when $x$ is negative, since the root cannot return a negative value.

Square both sides and gather terms.

$2x + 1 = x^2 - 2x + 1 \quad\Longrightarrow\quad x^2 - 4x = 0$

So $x(x - 4) = 0$, giving candidates $x = 0$ and $x = 4$. Checking, $x = 4$ gives $\sqrt{9} = 3$ and $4 - 1 = 3$, but $x = 0$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 4$ is a solution.

Take the square root of both sides, keeping both signs, and simplify the radical.

$x = \pm\sqrt{20} = \pm\sqrt{4 \cdot 5} = \pm 2\sqrt{5}$

Giving only the positive root would lose the solution $x = -2\sqrt{5}$.

Square both sides and form a quadratic.

$x = x^2 - 12x + 36 \quad\Longrightarrow\quad x^2 - 13x + 36 = 0$

Factoring gives $(x - 4)(x - 9) = 0$, so the candidates are $x = 4$ and $x = 9$. Checking, $x = 9$ gives $\sqrt{9} = 3 = 9 - 6$, but $x = 4$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 9$ survives.

A principal square root is never negative, so no input makes it equal $-3$.

$\sqrt{x} \ge 0 > -3$

Squaring would suggest $x = 9$, but $\sqrt{9} = 3 \ne -3$, so that candidate is extraneous and there is no real solution.

Take the square root of both sides, remembering the two signs.

$\lvert x - 3 \rvert = 4 \quad\Longrightarrow\quad x - 3 = \pm 4$

So $x = 3 + 4 = 7$ or $x = 3 - 4 = -1$. Writing only $x = \pm 4$ forgets the $-3$ inside.

The exponent $\tfrac{1}{2}$ is the square root, so square both sides.

$x^{1/2} = \sqrt{x} = 4 \quad\Longrightarrow\quad x = 16$

Check: $16^{1/2} = 4$.

If $a = b$ then $a^2 = b^2$, so every genuine solution survives the squaring.

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

Squaring may also add extra candidates, so it does not keep exactly the original solutions; the safe guarantee is only that none are lost.

Square both sides and form a quadratic.

$5x - 4 = x^2 \quad\Longrightarrow\quad x^2 - 5x + 4 = 0$

Factoring gives $(x - 1)(x - 4) = 0$. Checking, $x = 1$ gives $\sqrt{1} = 1$ and $x = 4$ gives $\sqrt{16} = 4$; both match, so both are solutions.

Square both sides, expanding the binomial in full.

$3x + 10 = x^2 + 4x + 4 \quad\Longrightarrow\quad x^2 + x - 6 = 0$

Factoring gives $(x + 3)(x - 2) = 0$, so the candidates are $x = -3$ and $x = 2$. Checking, $x = 2$ gives $\sqrt{16} = 4 = 2 + 2$, but $x = -3$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 2$ works.