To solve the equation [tex]\(\sqrt{3 - 3x} = 3 + \sqrt{3x + 2}\)[/tex], we will follow these algebraic steps:
1. Isolate one of the square roots:
[tex]\[
\sqrt{3 - 3x} = 3 + \sqrt{3x + 2}
\][/tex]
To isolate the square root on the left, we leave it as it is for now.
2. Square both sides to eliminate the square root:
[tex]\[
(\sqrt{3 - 3x})^2 = (3 + \sqrt{3x + 2})^2
\][/tex]
Simplifying both sides, we get:
[tex]\[
3 - 3x = 9 + 6\sqrt{3x + 2} + (3x + 2)
\][/tex]
3. Combine like terms:
[tex]\[
3 - 3x = 9 + 6\sqrt{3x + 2} + 3x + 2
\][/tex]
Simplifying the right side further:
[tex]\[
3 - 3x = 11 + 6\sqrt{3x + 2} + 3x
\][/tex]
4. Isolate the term involving the square root:
[tex]\[
3 - 3x - 11 - 3x = 6\sqrt{3x + 2}
\][/tex]
Combine the constants and linear terms on the left:
[tex]\[
-8 - 6x = 6\sqrt{3x + 2}
\][/tex]
5. Divide both sides by 6 to simplify:
[tex]\[
\frac{-8 - 6x}{6} = \sqrt{3x + 2}
\][/tex]
Simplifying the left side:
[tex]\[
-\frac{4 + 3x}{3} = \sqrt{3x + 2}
\][/tex]
6. Square both sides again to eliminate the remaining square root:
[tex]\[
\left(-\frac{4 + 3x}{3}\right)^2 = (3x + 2)
\][/tex]
Simplify each side:
[tex]\[
\left(\frac{4 + 3x}{3}\right)^2 = 3x + 2
\][/tex]
Expanding the square on the left:
[tex]\[
\frac{(4 + 3x)^2}{9} = 3x + 2
\][/tex]
7. Clear the fraction by multiplying each term by 9:
[tex]\[
(4 + 3x)^2 = 9(3x + 2)
\][/tex]
Expanding both sides:
[tex]\[
16 + 24x + 9x^2 = 27x + 18
\][/tex]
8. Combine like terms and set the equation to zero:
[tex]\[
9x^2 + 24x + 16 = 27x + 18
\][/tex]
Bring all the terms to one side:
[tex]\[
9x^2 + 24x + 16 - 27x - 18 = 0
\][/tex]
Simplifying:
[tex]\[
9x^2 - 3x - 2 = 0
\][/tex]
9. Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] to solve for [tex]\(x\)[/tex]:
Here, [tex]\(a = 9\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -2\)[/tex].
[tex]\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(9)(-2)}}{2(9)}
\][/tex]
Simplifying inside the square root:
[tex]\[
x = \frac{3 \pm \sqrt{9 + 72}}{18}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{81}}{18}
\][/tex]
[tex]\[
x = \frac{3 \pm 9}{18}
\][/tex]
Solving for the two cases:
[tex]\[
x = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3}
\][/tex]
[tex]\[
x = \frac{3 - 9}{18} = \frac{-6}{18} = -\frac{1}{3}
\][/tex]
10. Check both potential solutions in the original equation:
[tex]\[
\sqrt{3 - 3 \left( \frac{2}{3} \right)} = 3 + \sqrt{3 \left( \frac{2}{3} \right) + 2}
\][/tex]
[tex]\[
\sqrt{3 - 2} = 3 + \sqrt{2 + 2}
\][/tex]
[tex]\[
\sqrt{1} = 3 + \sqrt{4}
\][/tex]
[tex]\[
1 = 3 + 2
\][/tex]
This is not true, so [tex]\(x = \frac{2}{3}\)[/tex] is not a solution.
Similarly, for [tex]\(x = -\frac{1}{3}\)[/tex]:
[tex]\[
\sqrt{3 - 3 \left( -\frac{1}{3} \right)} = 3 + \sqrt{3 \left( -\frac{1}{3} \right) + 2}
\][/tex]
[tex]\[
\sqrt{3 + 1} = 3 + \sqrt{-1 + 2}
\][/tex]
[tex]\[
\sqrt{4} = 3 + \sqrt{1}
\][/tex]
[tex]\[
2 = 3 + 1
\][/tex]
This is also not true, so [tex]\(x = -\frac{1}{3}\)[/tex] is not a solution either.
Therefore, after thoroughly checking both potential solutions, we find that there are no valid solutions to the given equation. Thus, the solution set is empty.
[tex]\[
\boxed{\emptyset}
\][/tex]
