To solve the equation [tex]\(\sqrt{-3x - 2} = x + 2\)[/tex], we must identify any extraneous solutions. Let's start by solving the equation and then check each proposed solution.
### Step-by-Step Solution:
1. Square both sides of the equation to eliminate the square root:
[tex]\[
(\sqrt{-3x - 2})^2 = (x + 2)^2
\][/tex]
This simplifies to:
[tex]\[
-3x - 2 = (x + 2)^2
\][/tex]
2. Expand the right-hand side of the equation:
[tex]\[
(x + 2)^2 = x^2 + 4x + 4
\][/tex]
So, the equation becomes:
[tex]\[
-3x - 2 = x^2 + 4x + 4
\][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[
0 = x^2 + 4x + 4 + 3x + 2
\][/tex]
Combine like terms:
[tex]\[
x^2 + 7x + 6 = 0
\][/tex]
4. Factor the quadratic equation:
[tex]\[
x^2 + 7x + 6 = (x + 1)(x + 6) = 0
\][/tex]
This gives us two potential solutions:
[tex]\[
x + 1 = 0 \quad \text{or} \quad x + 6 = 0
\][/tex]
So:
[tex]\[
x = -1 \quad \text{or} \quad x = -6
\][/tex]
### Checking for Extraneous Solutions:
We need to check each solution to see if it satisfies the original equation [tex]\(\sqrt{-3x - 2} = x + 2\)[/tex]:
1. Check [tex]\(x = -1\)[/tex]:
[tex]\[
\sqrt{-3(-1) - 2} = \sqrt{3 - 2} = \sqrt{1} = 1
\][/tex]
[tex]\[
-1 + 2 = 1
\][/tex]
Since both sides equal 1, [tex]\(x = -1\)[/tex] is a valid solution.
2. Check [tex]\(x = -6\)[/tex]:
[tex]\[
\sqrt{-3(-6) - 2} = \sqrt{18 - 2} = \sqrt{16} = 4
\][/tex]
[tex]\[
-6 + 2 = -4
\][/tex]
Since [tex]\(\sqrt{16} = 4\)[/tex] does not equal [tex]\(-4\)[/tex], [tex]\(x = -6\)[/tex] is an extraneous solution.
Given the original equation and the checks that we performed, the extraneous solution is:
[tex]\[
\boxed{x = -6}
\][/tex]
