Let's solve the equation [tex]\(\sqrt{3x + 3} - 1 = x\)[/tex].
1. Step 1: Isolate the square root term.
[tex]\[
\sqrt{3x + 3} - 1 = x
\][/tex]
Add 1 to both sides:
[tex]\[
\sqrt{3x + 3} = x + 1
\][/tex]
2. Step 2: Square both sides to eliminate the square root.
[tex]\[
(\sqrt{3x + 3})^2 = (x + 1)^2
\][/tex]
Simplify:
[tex]\[
3x + 3 = (x + 1)^2
\][/tex]
[tex]\[
3x + 3 = x^2 + 2x + 1
\][/tex]
3. Step 3: Rearrange the equation to set it to zero.
[tex]\[
3x + 3 = x^2 + 2x + 1
\][/tex]
Subtract [tex]\(3x + 3\)[/tex] from both sides:
[tex]\[
0 = x^2 + 2x + 1 - 3x - 3
\][/tex]
[tex]\[
0 = x^2 - x - 2
\][/tex]
4. Step 4: Solve the quadratic equation.
The quadratic equation is:
[tex]\[
x^2 - x - 2 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x - 2)(x + 1) = 0
\][/tex]
Set each factor to zero:
[tex]\[
x - 2 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 2 \quad \text{or} \quad x = -1
\][/tex]
5. Step 5: Verify the solutions.
Substitute [tex]\(x = 2\)[/tex] back into the original equation:
[tex]\[
\sqrt{3(2) + 3} - 1 = 2
\][/tex]
[tex]\[
\sqrt{6 + 3} - 1 = 2
\][/tex]
[tex]\[
\sqrt{9} - 1 = 2
\][/tex]
[tex]\[
3 - 1 = 2
\][/tex]
[tex]\[
2 = 2 \quad \text{(True)}
\][/tex]
Substitute [tex]\(x = -1\)[/tex] back into the original equation:
[tex]\[
\sqrt{3(-1) + 3} - 1 = -1
\][/tex]
[tex]\[
\sqrt{-3 + 3} - 1 = -1
\][/tex]
[tex]\[
\sqrt{0} - 1 = -1
\][/tex]
[tex]\[
0 - 1 = -1
\][/tex]
[tex]\[
-1 = -1 \quad \text{(True)}
\][/tex]
Both solutions satisfy the original equation.
Therefore, the correct answer is:
A. [tex]\(-1\)[/tex] and [tex]\(2\)[/tex]
