Solve $\sqrt{x}=-\sqrt{x+5}+5$ for $x$.
The equation to solve is \(\sqrt{x} = -\sqrt{x+5} + 5\). First, we identify the domain of the variable \(x\). For the square root \(\sqrt{x}\) to be a real number, we must have \(x \geq 0\). Similarly, for \(\sqrt{x+5}\) to be a real number, we need \(x+5 \geq 0\), which implies \(x \geq -5\). Combining these, the valid domain for \(x\) is \(x \geq 0\). We can rewrite the original equation to isolate the radical terms: \(\sqrt{x} + \sqrt{x+5} = 5\).
To eliminate the square roots, we square both sides of the equation \(\sqrt{x} + \sqrt{x+5} = 5\):
\[(\sqrt{x} + \sqrt{x+5})^2 = 5^2\]
Expanding the left side using the identity \((a+b)^2 = a^2 + 2ab + b^2\):
\[x + 2\sqrt{x(x+5)} + (x+5) = 25\]
\[2x + 5 + 2\sqrt{x^2+5x} = 25\]
Now, we isolate the remaining radical term by subtracting \(2x + 5\) from both sides:
\[2\sqrt{x^2+5x} = 25 - 2x - 5\]
\[2\sqrt{x^2+5x} = 20 - 2x\]
Dividing the entire equation by 2, we obtain:
\[\sqrt{x^2+5x} = 10 - x\]
Note that for the left side to be equal to the right side, we must have \(10 - x \geq 0\), so \(x \leq 10\).
We square both sides again to remove the remaining radical:
\[(\sqrt{x^2+5x})^2 = (10 - x)^2\]
\[x^2 + 5x = 100 - 20x + x^2\]
Subtracting \(x^2\) from both sides simplifies the equation to a linear form:
\[5x = 100 - 20x\]
Adding \(20x\) to both sides, we get:
\[25x = 100\]
\[x = 4\]
Checking our condition \(x \leq 10\), we see that \(4 \leq 10\) is satisfied.
Finally, we must verify the solution in the original equation \(\sqrt{x} = -\sqrt{x+5} + 5\). Substituting \(x = 4\):
\[\sqrt{4} = -\sqrt{4+5} + 5\]
\[2 = -\sqrt{9} + 5\]
\[2 = -3 + 5\]
\[2 = 2\]
The solution holds true. Therefore, the solution to the equation is \(x=4\). The option corresponding to this is the third choice. Thus, the solution is \(\boxed{4}\).
This type of algebraic equation solving is essential when dealing with physical models involving square root relationships, such as calculating the time taken for an object to reach a specific point under certain force constraints, or when determining distances in geometry. For instance, if you are calculating the point where two sensor signals, whose strengths decrease with the square root of distance, balance out, you are essentially solving equations similar to the one we just worked through. Ensuring valid domains and verifying solutions are critical steps in any real-world engineering or physical calculation to avoid impossible results.