Radical Functions: Additional Practice (Set 1)

Square both sides and collect terms.

$3x - 2 = x^2 - 4x + 4 \;\Rightarrow\; x^2 - 7x + 6 = 0 \;\Rightarrow\; (x - 1)(x - 6) = 0$

Testing, $x = 6$ gives $\sqrt{16} = 4 = 6 - 2$, but $x = 1$ gives $\sqrt{1} = 1 \ne -1$. Only $x = 6$ survives.

Set the radicand nonnegative and solve.

$2x + 8 \ge 0 \;\Rightarrow\; 2x \ge -8 \;\Rightarrow\; x \ge -4$

Square both sides and collect terms.

$x + 9 = x^2 + 6x + 9 \;\Rightarrow\; x^2 + 5x = 0 \;\Rightarrow\; x(x + 5) = 0$

Testing, $x = 0$ gives $\sqrt{9} = 3 = 0 + 3$, but $x = -5$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 0$ survives.

Square both sides to remove both roots at once.

$4x + 5 = 6x - 3 \;\Rightarrow\; 8 = 2x \;\Rightarrow\; x = 4$

Checking, both sides equal $\sqrt{21}$, so $x = 4$ works.

The corner is at $(3, 2)$, with $a = 1 > 0$.

$x - 3 \ge 0 \Rightarrow x \ge 3, \qquad y \ge k = 2$

The domain starts at $x = 3$ and the range starts at $y = 2$.

Square both sides and factor.

$x - 1 = x^2 - 14x + 49 \;\Rightarrow\; x^2 - 15x + 50 = 0 \;\Rightarrow\; (x - 5)(x - 10) = 0$

Testing, $x = 10$ gives $\sqrt{9} = 3 = 10 - 7$, but $x = 5$ gives $\sqrt{4} = 2 \ne -2$. Only $x = 10$ survives.

The root is real only when the radicand is nonnegative, that is $x \ge 4$.

$x = 1 \;\Rightarrow\; \sqrt{1 - 4} = \sqrt{-3} \text{ is not real}$

The other values are at least $4$, and $x = 4$ gives $\sqrt{0} = 0$.

Square both sides and factor.

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

The right side $-x$ must be nonnegative, so $x \le 0$. Testing, $x = -1$ gives $\sqrt{1} = 1 = -(-1)$, but $x = 2$ gives $\sqrt{4} = 2 \ne -2$. Only $x = -1$ survives.

Work from the inside out.

$\sqrt{16} = 4, \qquad \sqrt{4} = 2$

So $\sqrt{\sqrt{16}} = 2$.

A shift right by $6$ replaces $x$ with $x - 6$, and a shift down by $1$ subtracts $1$ outside.

$y = \sqrt{x - 6} - 1$

Square both sides and solve for $x$.

$x^2 - 5 = 4 \;\Rightarrow\; x^2 = 9 \;\Rightarrow\; x = \pm 3$

Both values give $\sqrt{9 - 5} = \sqrt{4} = 2$, since the radicand depends only on $x^2$, so both check.

Substitute $A = 144$ and take the root.

$s(144) = \sqrt{144} = 12$