Solve the radical equation \( \sqrt{x + 1} = 3x - 2 \) and verify if the solution is a point on the line \( y = 2x + 1 \)

Step 1 ：First, we isolate the radical. We already have \( \sqrt{x + 1} = 3x - 2 \)

Step 2 ：Next, we square both sides to get rid of the square root, yielding \(x + 1 = (3x - 2)^2\)

Step 3 ：This simplifies to \(x + 1 = 9x^2 - 12x + 4\)

Step 4 ：Rearranging, we get \(9x^2 - 13x + 3 = 0\)

Step 5 ：This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Here we use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 6 ：Substituting, we get \(x = \frac{13 \pm \sqrt{169 - 108}}{18} = \frac{13 \pm \sqrt{61}}{18}\)

Step 7 ：So, we have two potential solutions: \(x = \frac{13 + \sqrt{61}}{18}\) or \(x = \frac{13 - \sqrt{61}}{18}\)

Step 8 ：However, we must check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

Step 9 ：After checking, we find that only \(x = \frac{13 - \sqrt{61}}{18}\) is a valid solution to the original equation.

Step 10 ：Finally, we check if this point is on the line by substituting \(x = \frac{13 - \sqrt{61}}{18}\) into \(y = 2x + 1\)

Step 11 ：Doing this, we find that \(y = 2\left(\frac{13 - \sqrt{61}}{18}\right) + 1\)