To determine if there are any extraneous solutions for the equation \(-3 - 2\sqrt{-x} = 2 - \sqrt{-x}\), we will follow a systematic approach.
Step 1: Rearrange the Equation
Start by isolating the radical terms on one side:
\[ -3 - 2\sqrt{-x} = 2 - \sqrt{-x} \]
Add \(\sqrt{-x}\) to both sides:
\[ -3 - 2\sqrt{-x} + \sqrt{-x} = 2 \]
This simplifies to:
\[ -3 - \sqrt{-x} = 2 \]
Next, add 3 to both sides:
\[ -\sqrt{-x} = 5 \]
Step 2: Solve for \(\sqrt{-x}\)
Multiply both sides by -1:
\[ \sqrt{-x} = -5 \]
Step 3: Square Both Sides
We will now square both sides to eliminate the square root:
\[ (-x) = (-5)^2 \]
This leads to:
\[ -x = 25 \]
Next, solve for \(x\):
\[ x = -25 \]
Step 4: Check for Extraneous Solutions
Now that we have \(x = -25\), we need to check whether this solution satisfies the original equation. Substitute \(x = -25\) back into the original equation:
\[ -3 - 2\sqrt{-(-25)} = 2 - \sqrt{-(-25)} \]
Calculate \(-(-25)\):
\[ -3 - 2\sqrt{25} = 2 - \sqrt{25} \]
Since \(\sqrt{25} = 5\), we substitute:
\[ -3 - 2(5) = 2 - 5 \]
This becomes:
\[ -3 - 10 = 2 - 5 \]
Calculating both sides:
Left side:
\[ -3 - 10 = -13 \]
Right side:
\[ 2 - 5 = -3 \]
Step 5: Compare Results
The left side gives \(-13\) and the right side gives \(-3\). Since \(-13 \neq -3\), \(x = -25\) does not satisfy the original equation.
Conclusion
Thus, the solution \(x = -25\) is extraneous because it does not satisfy the original equation. Therefore, there are extraneous solutions for the equation \( -3 - 2\sqrt{-x} = 2 - \sqrt{-x} \).