Question

which statement about \\(\sqrt{x - 5} - \sqrt{x} = 5\\) is true?
\\(\bigcirc\\) \\(x = -3\\) is a true solution.
\\(\bigcirc\\) \\(x = -3\\) is an extraneous solution.
\\(\bigcirc\\) \\(x = 9\\) is a true solution.
\\(\bigcirc\\) \\(x = 9\\) is an extraneous solution.

Explanation:

Step1: Analyze Domain

For $\sqrt{x - 5}$ and $\sqrt{x}$ to be real, $x - 5\geq0$ and $x\geq0$. So $x\geq5$. Thus, $x = - 3$ is not in the domain, and $x = 9$ is in the domain.

Step2: Test $x = 9$

Substitute $x = 9$ into the equation: $\sqrt{9 - 5}-\sqrt{9}=\sqrt{4}-3 = 2 - 3=-1
eq5$. Wait, maybe I made a mistake. Wait, the equation is $\sqrt{x - 5}-\sqrt{x}=5$? Wait, let's re - check. If $x = 9$, $\sqrt{9 - 5}=2$, $\sqrt{9}=3$, so $2-3=-1
eq5$. Wait, maybe the equation is $\sqrt{x - 5}+\sqrt{x}=5$? No, the original equation is $\sqrt{x - 5}-\sqrt{x}=5$. Let's check the domain again. For $\sqrt{x-5}$, $x\geq5$, for $\sqrt{x}$, $x\geq0$. So domain is $x\geq5$. Now, let's consider the left - hand side (LHS) of the equation $y=\sqrt{x - 5}-\sqrt{x}$. As $x$ increases, $\sqrt{x-5}$ and $\sqrt{x}$ both increase, but the rate of increase of $\sqrt{x}$ is faster than that of $\sqrt{x - 5}$. So $y=\sqrt{x - 5}-\sqrt{x}=\frac{(x - 5)-x}{\sqrt{x - 5}+\sqrt{x}}=\frac{-5}{\sqrt{x - 5}+\sqrt{x}}<0$ for all $x\geq5$. But the right - hand side (RHS) is $5>0$. So there is no solution in the domain. But let's check the options. Wait, maybe the equation was written wrong. Wait, maybe it's $\sqrt{x + 5}-\sqrt{x}=5$? No, the original problem says $\sqrt{x - 5}-\sqrt{x}=5$. Wait, let's check $x=-3$: $\sqrt{-3 - 5}$ is not real, so $x = - 3$ is extraneous. For $x = 9$, we saw LHS is $-1
eq5$, so $x = 9$ is extraneous? Wait, no, the domain is $x\geq5$, so $x=-3$ is not in the domain (extraneous), and $x = 9$ is in the domain but does not satisfy the equation, so it's also extraneous? But the options: Let's re - evaluate. Wait, maybe the equation is $\sqrt{x + 5}-\sqrt{x}=5$? No, the user's equation is $\sqrt{x - 5}-\sqrt{x}=5$. Wait, let's check the options again. The first two options are about $x=-3$ (not in domain, so extraneous), the last two about $x = 9$ (in domain but doesn't satisfy, so extraneous? But the options:

Wait, maybe I misread the equation. Let's assume the equation is $\sqrt{x + 5}-\sqrt{x}=5$ by mistake. Then for $x = 9$: $\sqrt{14}-\sqrt{9}=\sqrt{14}-3\approx3.74 - 3 = 0.74
eq5$. No. Wait, maybe the equation is $\sqrt{x - 5}+\sqrt{x}=5$. Then for $x = 9$: $\sqrt{4}+\sqrt{9}=2 + 3 = 5$. Ah! Maybe there was a sign error in the original equation. If the equation is $\sqrt{x - 5}+\sqrt{x}=5$, then:

Step1: Domain

$x-5\geq0\Rightarrow x\geq5$ and $x\geq0$, so $x\geq5$.

Step2: Solve the equation $\sqrt{x - 5}+\sqrt{x}=5$

Isolate one square root: $\sqrt{x - 5}=5-\sqrt{x}$
Square both sides: $x - 5 = 25-10\sqrt{x}+x$
Simplify: $x - 5-x=25-10\sqrt{x}\Rightarrow - 5 - 25=-10\sqrt{x}\Rightarrow - 30=-10\sqrt{x}\Rightarrow\sqrt{x}=3\Rightarrow x = 9$

Step3: Check $x = 9$ in original equation

$\sqrt{9 - 5}+\sqrt{9}=\sqrt{4}+3 = 2 + 3 = 5$, which matches the RHS. So if the equation was $\sqrt{x - 5}+\sqrt{x}=5$, then $x = 9$ is a true solution. But the original equation is $\sqrt{x - 5}-\sqrt{x}=5$. There must be a typo. Given the options, let's assume the equation is $\sqrt{x - 5}+\sqrt{x}=5$. Then:

- For $x=-3$: Not in domain (extraneous).
- For $x = 9$: Satisfies the equation (true solution).

But according to the original equation as written ($\sqrt{x - 5}-\sqrt{x}=5$), LHS is negative, RHS is positive, no solution. But the options suggest that we consider the equation might have a typo. Given the options, the most probable is that the equation is $\sqrt{x - 5}+\sqrt{x}=5$, and $x = 9$ is a true solution. Wait, no, in the original equation, if we proceed with the given equation $\sqrt{x - 5}-\sqrt{x}=5$:

Domain: $x\geq5$.

Let's check $x = 9$: $\sqrt{9 - 5}-\sqrt{9}=2 - 3=-1
eq5$.

Check $x=-3$: Not in domain (extraneous).

But the options are:

1. $x=-3$ is a true solution: False (not in domain).

2. $x=-3$ is an extraneous solution: True (not in domain, so extraneous).

3. $x = 9$ is a true solution: False (doesn't satisfy).

4. $x = 9$ is an extraneous solution: False (it's in the domain but doesn't satisfy, so it's a solution that doesn't work, but extraneous usually refers to solutions introduced during squaring that are not in the domain. Wait, no: Extraneous solutions are solutions obtained during the solution process (like squaring) that do not satisfy the original equation, even if they are in the domain.

Wait, let's solve the equation $\sqrt{x - 5}-\sqrt{x}=5$:

Isolate $\sqrt{x - 5}=5+\sqrt{x}$

Square both sides: $x - 5 = 25 + 10\sqrt{x}+x$

Simplify: $x - 5-x=25 + 10\sqrt{x}\Rightarrow - 5-25 = 10\sqrt{x}\Rightarrow - 30 = 10\sqrt{x}\Rightarrow\sqrt{x}=-3$, which has no real solution. So the solution process gives no real solutions. But the options are about $x=-3$ and $x = 9$. There must be a mistake in the equation. Given that $x = 9$ is in the domain of a similar equation (with a plus sign), and the options, we'll proceed with the assumption that the equation is $\sqrt{x - 5}+\sqrt{x}=5$.

So, for $x=-3$: Not in domain (extraneous).

For $x = 9$: Satisfies the equation (true solution). But in the original equation, it's different. However, based on the options, the correct answer is that $x = 9$ is an extraneous solution? No, wait, no. Wait, maybe the original equation is $\sqrt{x + 5}-\sqrt{x}=5$. Let's check $x = 9$: $\sqrt{14}-3\approx3.74 - 3 = 0.74
eq5$. No.

Wait, I think the key is that for the equation $\sqrt{x - 5}-\sqrt{x}=5$, the left - hand side is always negative (since $\sqrt{x}>\sqrt{x - 5}$ for $x>5$), and the right - hand side is positive. So there is no solution. But the options are about $x=-3$ (extraneous, since not in domain) and $x = 9$ (in domain but doesn't satisfy, so extraneous). But the options:

The first option: $x=-3$ is a true solution: False (not in domain).

Second: $x=-3$ is an extraneous solution: True (not in domain, so extraneous).

Third: $x = 9$ is a true solution: False (doesn't satisfy).

Fourth: $x = 9$ is an extraneous solution: True (in domain but doesn't satisfy). But the options are given, and maybe the equation was written wrong. Given that, the most probable answer is that $x = 9$ is an extraneous solution? Wait, no. Wait, let's go back.

Wait, the original equation: $\sqrt{x - 5}-\sqrt{x}=5$.

Let's solve it properly:

1. Domain: $x-5\geq0\Rightarrow x\geq5$ and $x\geq0\Rightarrow x\geq5$.

2. Let's rewrite the equation as $\sqrt{x - 5}=\sqrt{x}+5$.

3. Square both sides: $x - 5=x + 10\sqrt{x}+25$.

4. Simplify: $x - 5-x=10\sqrt{x}+25\Rightarrow - 5-25 = 10\sqrt{x}\Rightarrow - 30 = 10\sqrt{x}\Rightarrow\sqrt{x}=-3$.

5. Since the square root of a real number is non - negative, $\sqrt{x}=-3$ has no real solution. So there are no real solutions. But the options are about $x=-3$ and $x = 9$.

Now, $x=-3$: Not in the domain (extraneous, because it was obtained if we didn't consider the domain, but actually, when we solved, we got no real solutions. But if we mistakenly solved without considering the domain first, and substituted $x=-3$, it's not in the domain, so extraneous.

$x = 9$: In the domain, but when we substitute, it doesn't satisfy the equation, so it's an extraneous solution (because it was not a solution of the original equation, even though it's in the domain).

But among the options, the statement "x = 9 is an extraneous solution" is true? Wait, no, when we solved the equation, we got no real solutions. But if we made a mistake in the solution process (like squaring without considering the domain properly), and thought $x = 9$ was a solution, but it's not, so it's extraneous.

But let's check the options again. The fourth option is "x = 9 is an extraneous solution".

Answer:

D. $x = 9$ is an extraneous solution (assuming the options are labeled as A: $x=-3$ is a true solution, B: $x=-3$ is an extraneous solution, C: $x = 9$ is a true solution, D: $x = 9$ is an extraneous solution)