Raising Equations to Powers: Additional Practice (Set 2)

The factor $(x - 2)$ is zero at $x = 2$, so multiplying by it makes both sides zero there, whether or not $x = 2$ solved the original.

$\text{at } x = 2\!: \ (x - 2)\cdot(\text{anything}) = 0 = 0$

So $x = 2$ may satisfy the new equation without satisfying the original; it is an introduced, possibly extraneous, solution to be checked.

Square both sides, keeping the surviving radical.

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

Squaring again gives $x^2 + 4x + 4 = 4x$, so $x^2 + 4 = 0$. No real number satisfies this, so the equation has no real solution.

Square both sides; both roots vanish at once.

$3x - 2 = x + 6 \quad\Longrightarrow\quad 2x = 8 \quad\Longrightarrow\quad x = 4$

Check: $\sqrt{3 \cdot 4 - 2} = \sqrt{10}$ and $\sqrt{4 + 6} = \sqrt{10}$, so $x = 4$.

Squaring is not reversible, so it can add candidates the original never had.

$x = 3 \quad\Longrightarrow\quad x^2 = 9 \quad\Longrightarrow\quad x = \pm 3$

Adding a constant, subtracting a term, and multiplying by a nonzero constant are all reversible, so they change nothing about the solution set.

Square both sides, expanding the binomial.

$2x + 5 = x^2 + 2x + 1 \quad\Longrightarrow\quad x^2 = 4 \quad\Longrightarrow\quad x = \pm 2$

Checking, $x = 2$ gives $\sqrt{9} = 3 = 2 + 1$, but $x = -2$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 2$ is a solution.

Take the square root of both sides, keeping both signs.

$\lvert 2x - 1 \rvert = 5 \quad\Longrightarrow\quad 2x - 1 = \pm 5$

So $2x = 6$ gives $x = 3$, and $2x = -4$ gives $x = -2$. Both are solutions.

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

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

Since $s \ge 0$, only $s = 2$ is allowed, so $x = s^2 = 4$. Check: $4 + \sqrt{4} = 4 + 2 = 6$.

Square both sides to clear the root.

$9x = x^2 + 4x + 4 \quad\Longrightarrow\quad x^2 - 5x + 4 = 0 \quad\Longrightarrow\quad (x - 1)(x - 4) = 0$

Checking, $x = 1$ gives $3\sqrt{1} = 3 = 1 + 2$, and $x = 4$ gives $3\sqrt{4} = 6 = 4 + 2$. Both check, so both are solutions.

The original equation requires the root's value $B$ to be nonnegative, because a principal root is never negative.

$\sqrt{A} = B \ \text{ requires } \ B \ge 0$

Squaring drops that requirement, which is exactly how a candidate with $B < 0$ becomes extraneous. So the step is safe precisely when $B \ge 0$.

Square both sides and form a quadratic.

$x + 20 = x^2 \quad\Longrightarrow\quad x^2 - x - 20 = 0 \quad\Longrightarrow\quad (x - 5)(x + 4) = 0$

The candidates are $x = 5$ and $x = -4$. Checking, $x = 5$ gives $\sqrt{25} = 5$, but $x = -4$ gives $\sqrt{16} = 4 \ne -4$. Only $x = 5$ works.

Square both sides, expanding the binomial.

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

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

Square both sides to get a quadratic.

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

The candidates are $x = 0$ and $x = 1$. Checking, $x = 0$ gives $\sqrt{0} = 0 = -0$, but $x = 1$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 0$ is a real solution.