Raising Equations to Powers: Practice

Isolate one radical, then square.

$\sqrt{2x + 7} = 1 + \sqrt{x + 3} \quad\Longrightarrow\quad 2x + 7 = x + 4 + 2\sqrt{x + 3}$

This simplifies to $x + 3 = 2\sqrt{x + 3}$; squaring again gives $(x+3)^2 = 4(x+3)$, so $(x+3)(x-1) = 0$ and the candidates are $x = -3$ and $x = 1$. Both check: at $x = -3$, $\sqrt{1} - \sqrt{0} = 1$, and at $x = 1$, $\sqrt{9} - \sqrt{4} = 1$. Both are solutions.

Isolate one radical before squaring.

$\sqrt{x + 5} = 5 - \sqrt{x} \quad\Longrightarrow\quad x + 5 = 25 - 10\sqrt{x} + x$

This leaves $10\sqrt{x} = 20$, so $\sqrt{x} = 2$ and $x = 4$. Check: $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$.

Write $x^{2/3} = \left(x^{1/3}\right)^2$ and let $u = x^{1/3}$.

$u^2 = 9 \quad\Longrightarrow\quad u = \pm 3 \quad\Longrightarrow\quad x = (\pm 3)^3 = \pm 27$

Both check, since $(\pm 27)^{2/3} = (\pm 3)^2 = 9$. The even numerator produces the two signs.

Raise both sides to the reciprocal power $\tfrac{2}{3}$.

$x = 27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$

The odd numerator gives a single solution, and $x \ge 0$ rules out any negative value. Check: $9^{3/2} = 3^3 = 27$.

Take the square root of both sides, which introduces two sign cases.

$\lvert x - 4 \rvert = \lvert 2x + 1 \rvert \quad\Longrightarrow\quad x - 4 = 2x + 1 \ \text{ or } \ x - 4 = -(2x + 1)$

The first gives $x = -5$; the second gives $3x = 3$, so $x = 1$. Both are solutions.

Square both sides, then solve for $x$ keeping both signs.

$x^2 - 5 = 4 \quad\Longrightarrow\quad x^2 = 9 \quad\Longrightarrow\quad x = \pm 3$

Both check: $\sqrt{9 - 5} = \sqrt{4} = 2$ for $x = 3$ and for $x = -3$, since $x$ appears only as $x^2$.

Square both sides and collect into a quadratic.

$x + 9 = x^2 + 6x + 9 \quad\Longrightarrow\quad x^2 + 5x = 0$

So $x(x + 5) = 0$, giving candidates $x = 0$ and $x = -5$. Checking, $x = 0$ gives $\sqrt{9} = 3 = 0 + 3$, but $x = -5$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 0$ works.

Square both sides and form a quadratic.

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

So $x(x - 5) = 0$, with candidates $x = 0$ and $x = 5$. Checking, $x = 5$ gives $\sqrt{9} = 3 = 5 - 2$, but $x = 0$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 5$ is a solution.

Square both sides and solve.

$x = x^2 - 4x + 4 \quad\Longrightarrow\quad x^2 - 5x + 4 = 0 \quad\Longrightarrow\quad (x - 1)(x - 4) = 0$

The candidates are $x = 1$ and $x = 4$. Checking, $x = 4$ gives $\sqrt{4} = 2 = 4 - 2$, but $x = 1$ gives $\sqrt{1} = 1 \ne -1$. Exactly one solution survives.

Move everything to one side and factor rather than dividing by $x$.

$x^2 - 4x = 0 \quad\Longrightarrow\quad x(x - 4) = 0$

The solutions are $x = 0$ and $x = 4$. Substituting $x = 2$ gives $4 \ne 8$, so $x = 2$ is not a solution.

Isolate the radical first.

$\sqrt{x - 3} = 2 - 5 = -3$

A principal square root cannot equal a negative number, so there is no real solution. Squaring would give $x = 12$, but $\sqrt{9} = 3 \ne -3$, so that candidate is extraneous.

Since $x^2 \ge 0$, taking the square root of both sides gives $x^2 = 4$ (the value $-4$ is impossible).

$x^4 = 16 \quad\Longrightarrow\quad x^2 = 4 \quad\Longrightarrow\quad x = \pm 2$

Both check, since $(\pm 2)^4 = 16$.