Radical Functions: Additional Practice (Set 2)

Square both sides and collect terms.

$x + 7 = x^2 + 2x + 1 \;\Rightarrow\; x^2 + x - 6 = 0 \;\Rightarrow\; (x + 3)(x - 2) = 0$

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

Isolate the radical, then square.

$\sqrt{x} = x - 2 \;\Rightarrow\; x = x^2 - 4x + 4 \;\Rightarrow\; x^2 - 5x + 4 = 0 \;\Rightarrow\; (x - 1)(x - 4) = 0$

Testing, $x = 4$ gives $\sqrt{4} + 2 = 4$, but $x = 1$ gives $\sqrt{1} + 2 = 3 \ne 1$. Only $x = 4$ survives.

Square both sides and factor.

$3x + 1 = x^2 - 6x + 9 \;\Rightarrow\; x^2 - 9x + 8 = 0 \;\Rightarrow\; (x - 1)(x - 8) = 0$

Testing, $x = 8$ gives $\sqrt{25} = 5 = 8 - 3$, but $x = 1$ gives $\sqrt{4} = 2 \ne -2$. Exactly one solution, $x = 8$, survives.

Take the fourth root first, then cube the result.

$16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$

The radicand must be nonnegative, giving a quadratic inequality.

$x^2 - 9 \ge 0 \;\Rightarrow\; x^2 \ge 9 \;\Rightarrow\; x \le -3 \text{ or } x \ge 3$

Between $-3$ and $3$ the radicand is negative, so those inputs are excluded.

Square both sides and factor.

$2x + 1 = x^2 - 14x + 49 \;\Rightarrow\; x^2 - 16x + 48 = 0 \;\Rightarrow\; (x - 4)(x - 12) = 0$

Testing, $x = 12$ gives $\sqrt{25} = 5 = 12 - 7$, but $x = 4$ gives $\sqrt{9} = 3 \ne -3$. Exactly one solution survives.

The corner is at $(h, k) = (1, 3)$, and $a = -2 < 0$ turns the curve downward.

$a < 0 \;\Rightarrow\; y \le k = 3$

Starting at the corner height $3$, the outputs only decrease.

The cube root accepts any real input, but the square root requires $x \ge 0$, and both conditions must hold.

$x \ge 0$

The stricter square-root condition sets the domain.

The principal square root is never negative, so it cannot equal $-3$.

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

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

Isolate the radical, then square.

$\sqrt{x} = 4 \;\Rightarrow\; x = 16$

Checking, $\sqrt{16} + 1 = 4 + 1 = 5$.

Square both sides and collect terms.

$4x + 1 = 4x^2 - 4x + 1 \;\Rightarrow\; 4x^2 - 8x = 0 \;\Rightarrow\; 4x(x - 2) = 0$

Testing, $x = 2$ gives $\sqrt{9} = 3 = 2(2) - 1$, but $x = 0$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 2$ survives.

Compute each term separately.

$\left(\sqrt{9}\right)^2 = 9, \qquad \sqrt{9^2} = \sqrt{81} = 9$

Subtracting gives $9 - 9 = 0$.