We are given the equation:

\[
\sqrt{3\sqrt{x+1}} = \sqrt{3x - 5}
\]

### Step 1: Square both sides of the equation to eliminate the square roots.

\[
\left(\sqrt{3\sqrt{x+1}}\right)^2 = \left(\sqrt{3x - 5}\right)^2
\]

This simplifies to:

\[
3\sqrt{x+1} = 3x - 5
\]

### Step 2: Divide both sides by 3.

\[
\sqrt{x+1} = x - \frac{5}{3}
\]

### Step 3: Square both sides again to eliminate the remaining square root.

\[
\left(\sqrt{x+1}\right)^2 = \left(x - \frac{5}{3}\right)^2
\]

This simplifies to:

\[
x + 1 = \left(x - \frac{5}{3}\right)^2
\]

### Step 4: Expand the right-hand side.

\[
x + 1 = \left(x - \frac{5}{3}\right)^2 = x^2 - 2 \cdot x \cdot \frac{5}{3} + \left(\frac{5}{3}\right)^2
\]

\[
x + 1 = x^2 - \frac{10x}{3} + \frac{25}{9}
\]

### Step 5: Multiply the entire equation by 9 to eliminate fractions.

\[
9(x + 1) = 9\left(x^2 - \frac{10x}{3} + \frac{25}{9}\right)
\]

\[
9x + 9 = 9x^2 - 30x + 25
\]

### Step 6: Rearrange the equation.

\[
9x + 9 - 9x^2 + 30x - 25 = 0
\]

\[
-9x^2 + 39x - 16 = 0
\]

### Step 7: Solve the quadratic equation.

Multiply through by \(-1\) to simplify:

\[
9x^2 - 39x + 16 = 0
\]

Use the quadratic formula:

\[
x = \frac{-(-39) \pm \sqrt{(-39)^2 - 4(9)(16)}}{2(9)}
\]

\[
x = \frac{39 \pm \sqrt{1521 - 576}}{18}
\]

\[
x = \frac{39 \pm \sqrt{945}}{18}
\]

\[
x = \frac{39 \pm 3\sqrt{105}}{18}
\]

\[
x = \frac{13 \pm \sqrt{105}}{6}
\]

### Step 8: Approximate the values.

The approximate values for \(x\) are:

\[
x_1 \approx 3.99, \quad x_2 \approx 0.34
\]

### Step 9: Check for extraneous solutions.

Substitute these values back into the original equation to verify if they are valid.

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Do you have any questions or want a detailed explanation of any step?

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Here are some related problems you might find interesting:
1. Solve \(\sqrt{2x + 3} = x - 1\).
2. Solve \(\sqrt{4x - 7} = x - 2\).
3. Solve \(\sqrt{x + 5} = 3 - x\).
4. Solve \( \sqrt{5x - 4} = 2x - 3\).
5. Solve \( \sqrt{x - 1} = 2x - 3\).

**Tip:** Always check for extraneous solutions when solving equations involving square roots, as squaring both sides can introduce invalid solutions.

Explore how to solve the radical equation \sqrt{3\sqrt{x+1}} = \sqrt{3x-5} step-by-step. This problem delves into key algebraic concepts such as squaring both sides, solving quadratic equations, and checking for extraneous solutions. Ideal for students in Grades 9-11.

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