Question

determine for what real values of x, if any, the following statements are true.
a. \\(\sqrt{x} = 4\\) b. \\(\sqrt{x} = -4\\) c. \\(\sqrt{-x} = 4\\)
d. \\(\sqrt{-x} = -4\\) e. \\(\sqrt{x} > 0\\) f. \\(\sqrt{x} < 0\\)

a. select the correct choice below and, if necessary, fill in the answer box within your choice.
\\(\bigcirc\\) a. the solution is \\(x = \square\\).
\\(\bigcirc\\) b. the solution is \\(\\{x \mid x < \square\\}\\).
\\(\bigcirc\\) c. the solution is \\(\\{x \mid x > \square\\}\\).
\\(\bigcirc\\) d. there is no real solution.

b. select the correct choice below and, if necessary, fill in the answer box within your choice.
\\(\bigcirc\\) a. the solution is \\(x = \square\\).
\\(\bigcirc\\) b. the solution is \\(\\{x \mid x < \square\\}\\).
\\(\bigcirc\\) c. the solution is \\(\\{x \mid x > \square\\}\\).
\\(\bigcirc\\) d. there is no real solution.

c. select the correct choice below and, if necessary, fill in the answer box within your choice.
\\(\bigcirc\\) a. the solution is \\(x = \square\\).
\\(\bigcirc\\) b. the solution is \\(\\{x \mid x < \square\\}\\).
\\(\bigcirc\\) c. the solution is \\(\\{x \mid x > \square\\}\\).
\\(\bigcirc\\) d. there is no real solution.

d. select the correct choice below and, if necessary, fill in the answer box within your choice.
\\(\bigcirc\\) a. the solution is \\(x = \square\\).
\\(\bigcirc\\) b. the solution is \\(\\{x \mid x < \square\\}\\).
\\(\bigcirc\\) c. the solution is \\(\\{x \mid x > \square\\}\\).
\\(\bigcirc\\) d. there is no real solution.

Explanation:

Part a

Step1: Recall square root property

The principal square root $\sqrt{x}$ is non - negative, and if $\sqrt{x}=a$ ($a\geq0$), then $x = a^{2}$.
Given $\sqrt{x}=4$, we square both sides of the equation.

Step2: Solve for x

Squaring both sides, we get $x = 4^{2}=16$. Also, we need to check the domain of the square root function. The domain of $y = \sqrt{x}$ is $x\geq0$, and when $x = 16$, $\sqrt{16}=4$, which is valid.

Step1: Analyze the square root's range

The principal square root function $y=\sqrt{x}$ has a range of $y\geq0$. That is, $\sqrt{x}$ is always greater than or equal to 0 for all $x$ in its domain ($x\geq0$).

Step2: Determine the solution

The equation $\sqrt{x}=- 4$ is asking for a non - negative number (since $\sqrt{x}\geq0$) to be equal to a negative number (-4). This is impossible.

Step1: Recall square root property for $\sqrt{-x}$

If $\sqrt{-x}=4$, first, the expression under the square root, $-x$, must be non - negative (because the domain of $y = \sqrt{u}$ is $u\geq0$), so $-x\geq0\Rightarrow x\leq0$. Also, if $\sqrt{-x}=4$, we square both sides.

Step2: Solve for x

Squaring both sides of $\sqrt{-x}=4$, we get $-x=4^{2}=16$, then $x=- 16$. We check the domain: when $x = - 16$, $-x=16\geq0$, and $\sqrt{16}=4$, so the solution is valid.

Answer:

A. The solution is $x=\boldsymbol{16}$.

Part b