An algebra problem by A Former Brilliant Member

$\begin{aligned} \frac {x-\sqrt{x+1}}{x+\sqrt{x+1}} & = \frac {11}5 & \small \color{#3D99F6} \text{Let } u^2 = x + 1, \ x = u^2-1 \\ \frac {u^2-1-u}{u^2-1+u} & = \frac {11}5 \\ 5(u^2-u-1) & = 11(u^2+u-1) \\ 6u^2 + 16 u - 6 & = 0 \\ 3u^2 + 8u - 3 & = 0 \\ (3u-1)(u+3) & = 0 \end{aligned}$

$\implies \begin{cases} u = \sqrt{x+1} = \frac 13 & \implies x = - \frac 89 & \implies \frac {- \frac 89-\sqrt{- \frac 89+1}}{- \frac 89+\sqrt{- \frac 89+1}} & \color{#3D99F6} = \frac {11}5 \quad \small \text{Accepted} \\ u = \sqrt{x+1} = -3 & \implies x = 8 & \implies \frac {8-\sqrt{8+1}}{8 + \sqrt {8+1}} & \color{#D61F06} \ne \frac {11}5 \quad \small \text{Rejected} \end{cases}$

Therefore, all the values of solution $x$ is $\boxed{-\frac 89}$ .

Rather cheesy, but it works too(of course if the question was open the other solutions are needed): - 8 gives a root of a negative, which is not allowed in this case I assume, since we are talking about x and not z. Then 8 gives a problem too: it doesn't lead to 11/5. Conclusion: - 8/9 is the only possibility left.

Relevant wiki: Quadratic Equations - Problem Solving