To solve the equation \(\sqrt{3x + 3} - 1 = x\), let's follow a step-by-step approach:
1. Isolate the square root term:
[tex]\[
\sqrt{3x + 3} = x + 1
\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[
(\sqrt{3x + 3})^2 = (x + 1)^2
\][/tex]
Simplifying both sides, we get:
[tex]\[
3x + 3 = (x + 1)^2
\][/tex]
Expanding the right side:
[tex]\[
3x + 3 = x^2 + 2x + 1
\][/tex]
3. Move all terms to one side to form a quadratic equation:
[tex]\[
3x + 3 - 3x = x^2 + 2x + 1 - 3x - 3
\][/tex]
Simplifying, we get:
[tex]\[
0 = x^2 - x - 2
\][/tex]
Rearranging, we have:
[tex]\[
x^2 - x - 2 = 0
\][/tex]
4. Factorize the quadratic equation:
We look for two numbers that multiply to \(-2\) and add up to \(-1\). These numbers are \(-2\) and \(1\). Thus, the quadratic can be factored as:
[tex]\[
(x - 2)(x + 1) = 0
\][/tex]
5. Find the roots of the factored equation:
Setting each factor to zero gives:
[tex]\[
x - 2 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Solving these, we find:
[tex]\[
x = 2 \quad \text{or} \quad x = -1
\][/tex]
6. Check for extraneous solutions:
- For \(x = 2\):
[tex]\[
\sqrt{3(2) + 3} - 1 = 2
\][/tex]
Simplifying the left-hand side:
[tex]\[
\sqrt{6 + 3} - 1 = 2 \implies \sqrt{9} - 1 = 2 \implies 3 - 1 = 2 \implies 2 = 2 \quad (\text{True})
\][/tex]
- For \(x = -1\):
[tex]\[
\sqrt{3(-1) + 3} - 1 = -1
\][/tex]
Simplifying the left-hand side:
[tex]\[
\sqrt{-3 + 3} - 1 = -1 \implies \sqrt{0} - 1 = -1 \implies 0 - 1 = -1 \implies -1 = -1 \quad (\text{True})
\][/tex]
Since both values satisfy the original equation, the solutions are \(x = -1\) and \(x = 2\).
Thus, the correct answer is:
#### A. -1 and 2
