Originally Posted by

**skeeter** $\displaystyle (\sqrt{x+1} + 2)^2 = (\sqrt{3x+1})^2$

$\displaystyle (x+1) + 4\sqrt{x+1} + 4 = 3x+1$

combine like terms on the left side ...

$\displaystyle 4\sqrt{x+1} + x + 5 = 3x + 1$

$\displaystyle 4\sqrt{x+1} = 2x - 4$

divide all terms by 2 ...

$\displaystyle 2\sqrt{x+1} = x - 2$

square again ...

$\displaystyle 4(x+1) = x^2 - 4x + 4$

$\displaystyle 4x + 4 = x^2 - 4x + 4$

$\displaystyle 0 = x^2 - 8x$

$\displaystyle 0 = x(x - 8)$

$\displaystyle x = 0$ ... makes no sense in the context of the problem

$\displaystyle x = 8$