Raising Equations to Powers: Additional Practice (Set 1)

Square both sides and collect into a quadratic.

$5x - 1 = x^2 + 2x + 1 \quad\Longrightarrow\quad x^2 - 3x + 2 = 0$

Factoring gives $(x - 1)(x - 2) = 0$. Checking, $x = 1$ gives $\sqrt{4} = 2 = 1 + 1$, and $x = 2$ gives $\sqrt{9} = 3 = 2 + 1$. Both check, so both are solutions.

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

$\sqrt{16x^2} = \sqrt{16}\,\sqrt{x^2} = 4\lvert x \rvert$

The absolute value is needed because $x$ could be negative, and a root never returns a negative value.

Write $x^{2/5} = \left(x^{1/5}\right)^2$ and let $u = x^{1/5}$.

$u^2 = 4 \quad\Longrightarrow\quad u = \pm 2 \quad\Longrightarrow\quad x = (\pm 2)^5 = \pm 32$

Both check, since $(\pm 32)^{2/5} = (\pm 2)^2 = 4$. The even numerator produces the two signs.

Let $s = \sqrt{x}$, so $x = s^2$ and $s \ge 0$.

$s^2 - 5s + 6 = 0 \quad\Longrightarrow\quad (s - 2)(s - 3) = 0$

So $s = 2$ or $s = 3$, giving $x = 4$ or $x = 9$. Both check: $4 - 5(2) + 6 = 0$ and $9 - 5(3) + 6 = 0$.

Square both sides and form a quadratic.

$x + 6 = x^2 \quad\Longrightarrow\quad x^2 - x - 6 = 0 \quad\Longrightarrow\quad (x - 3)(x + 2) = 0$

The candidates are $x = 3$ and $x = -2$. Checking, $x = 3$ gives $\sqrt{9} = 3$, but $x = -2$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 3$ works.

Test each candidate in the original equation.

$x = 6\!: \ \sqrt{4} = 2 = 6 - 4 \ \checkmark \qquad x = 3\!: \ \sqrt{1} = 1 \ne 3 - 4$

At $x = 3$ the right side is $-1$ while the root is $1$, so $x = 3$ is extraneous.

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

$x = 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$

The odd numerator gives a single solution. Check: $4^{3/2} = 2^3 = 8$.

The root of a square is the absolute value of the base.

$\sqrt{(3 - x)^2} = \lvert 3 - x \rvert$

This equals $3 - x$ only when $x \le 3$; for larger $x$ it equals $x - 3$, and the absolute value handles both.

Square both sides and gather terms.

$x + 5 = x^2 - 2x + 1 \quad\Longrightarrow\quad x^2 - 3x - 4 = 0 \quad\Longrightarrow\quad (x - 4)(x + 1) = 0$

The candidates are $x = 4$ and $x = -1$. Checking, $x = 4$ gives $\sqrt{9} = 3 = 4 - 1$, but $x = -1$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 4$ works.

A fourth power of a real number is never negative.

$x^4 \ge 0 > -16$

Since no real $x$ can make $x^4$ negative, the equation has no real solution.

Squaring a difference of two roots leaves a cross term with a radical, so isolate one root first.

$\sqrt{x - 1} = 1 + \sqrt{x - 4}$

Squaring this form leaves a single radical, which you then isolate and square a second time.

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

$x^2 + 9 = 25 \quad\Longrightarrow\quad x^2 = 16 \quad\Longrightarrow\quad x = \pm 4$

Both check, since $x$ appears only as $x^2$: $\sqrt{16 + 9} = \sqrt{25} = 5$ for $x = 4$ and for $x = -4$.