Let's solve the equation [tex]\( -4 \sqrt{x-3} = -12 \)[/tex] step by step and determine if the solution is extraneous.
### Step 1: Isolate the square root term
First, we want to isolate the square root term on one side of the equation:
[tex]\[ -4 \sqrt{x - 3} = -12 \][/tex]
Divide both sides by [tex]\(-4\)[/tex]:
[tex]\[ \sqrt{x - 3} = \frac{-12}{-4} \][/tex]
[tex]\[ \sqrt{x - 3} = 3 \][/tex]
### Step 2: Remove the square root
Next, square both sides of the equation to eliminate the square root:
[tex]\[ (\sqrt{x - 3})^2 = 3^2 \][/tex]
[tex]\[ x - 3 = 9 \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Now, solve for [tex]\( x \)[/tex]:
[tex]\[ x = 9 + 3 \][/tex]
[tex]\[ x = 12 \][/tex]
### Step 4: Check for an extraneous solution
An extraneous solution is a solution derived from an algebraic manipulation but does not satisfy the original equation. We substitute [tex]\( x = 12 \)[/tex] back into the original equation to verify:
[tex]\[ -4 \sqrt{12 - 3} = -4 \sqrt{9} = -4 \times 3 = -12 \][/tex]
Since substituting [tex]\( x = 12 \)[/tex] into the original equation results in [tex]\(-12\)[/tex], which matches the right-hand side of the original equation, it is not an extraneous solution.
### Conclusion
Therefore, the solution is:
[tex]\[ x = 12 \][/tex]
And it is not an extraneous solution. The correct option is:
[tex]\[ x = 12, \text{ solution is not extraneous} \][/tex]
