Question

This square root equation has 2 solutions. What is the extraneous solution?

x+12−−−−−√−6=x

Enter your answer in the box.

Answer 1

To solve the equation \( \sqrt{x + 12} - 6 = x \), we begin by isolating the square root:

\[
\sqrt{x + 12} = x + 6
\]

Next, we square both sides to eliminate the square root:

\[
(\sqrt{x + 12})^2 = (x + 6)^2
\]

This simplifies to:

\[
x + 12 = (x + 6)(x + 6) = x^2 + 12x + 36
\]

Now, we rearrange the equation to bring all terms to one side:

\[
0 = x^2 + 12x + 36 - x - 12
\]

Simplifying further, we combine like terms:

\[
0 = x^2 + 11x + 24
\]

Next, we can factor the quadratic equation:

\[
0 = (x + 3)(x + 8)
\]

Setting each factor equal to zero, we find the possible solutions:

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 8 = 0 \quad \Rightarrow \quad x = -8
\]

Now we have two potential solutions: \( x = -3 \) and \( x = -8 \). We need to check both solutions in the original equation to identify any extraneous solutions.

1. **Check \( x = -3 \)**:

\[
\sqrt{-3 + 12} - 6 = -3
\]
\[
\sqrt{9} - 6 = -3
\]
\[
3 - 6 = -3 \quad \Rightarrow \quad -3 = -3 \quad \text{(valid solution)}
\]

2. **Check \( x = -8 \)**:

\[
\sqrt{-8 + 12} - 6 = -8
\]
\[
\sqrt{4} - 6 = -8
\]
\[
2 - 6 = -8 \quad \Rightarrow \quad -4 = -8 \quad \text{(not valid)}
\]

Thus, the extraneous solution is:

\[
\boxed{-8}
\]