An algebra problem by Rakshit Joshi

let

$\sqrt{x} + \sqrt{x+2} = k$

squaring both sides we get

$\sqrt{(x)(x+2)} + x = \frac{(k^2-2)}{2}$

thus we may write the given equation as

$\frac{(k^2-2)}{2} +k = 3$

solving this we get $k=2$

$\sqrt{x} + \sqrt{x+2} = 2$

on squaring and simplifying we get $x=\frac{1}{4}$

$\begin{aligned} x + \sqrt x + \sqrt{x+2} + \sqrt{x^2+2x} & = 3 & \small \color{#3D99F6} \text{Let } x = u^2 \\ u^2 + u + \sqrt{u^2+2} + u \sqrt{u^2+2} & = 3 \\ \frac {3-u-u^2}{1+u} & = \sqrt{u^2+2} \\ \frac 3{1+u} - u & = \sqrt{u^2+2} & \small \color{#3D99F6} \text{Squaring both sides} \\ \frac 9{(1+u)^2} - \frac {6u}{1+u} + u^2 & = u^2 + 2 & \small \color{#3D99F6} \text{Multiplying both sides by }(1+u)^2 \\ 9 - 6u(1+u) & = 2(1+u)^2 \\ 9 - 6u - 6u^2 & = 2 + 4u + 2u^2 \\ 8u^2 + 10u -7 & = 0 \\ (2u-1)(4u+7) & = 0 \end{aligned}$

$\implies u = \begin{cases} \frac 12 & \implies x = \frac 14 \\ -\frac 74 & \implies x = \frac {49}{16} \end{cases}$

$\implies \begin{cases} x = \frac 14 & \implies \frac 14 + \sqrt{\frac 14} + \sqrt{\frac 14+2} + \sqrt{\frac 1{16}+\frac 12} \color{#3D99F6}= 3 & \color{#3D99F6} \text{accepted} \\ x = \frac {49}{16} & \implies \frac {49}{16} + \sqrt{\frac {49}{16}} + \sqrt{\frac {49}{16}+2} + \sqrt{\frac {2401}{256}+\frac {49}{8}} \color{#D61F06}=11 & \color{#D61F06} \text{rejected} \end{cases}$

Therefore, $x = \boxed{0.25}$

This problem is similar to @Pi Han Goh 's problem (you can check it here ). So, I think the answer is the same. By trial and error, it's true.. I don't know, whether it's good or not.. Hehe

Domain of LHS is x>0.We attempt to find a value of x such that the terms (of LHS) in the square roots become perfect squares.Let us assume that x=t² for some t>0.So, x+2=t²+2.Let t²+2=k² for some k>0.Hence, (k-t)(k+t)=2. Hence, k+t=2 and k-t=1 is a possible solution.So, we get t=1/2 which gives x=1/4. And, on substitution of x=1/4 in LHS it does become 3. Since the function (by function I mean, bring 3 on the LHS and the entire thing on LHS is the function) has positive derivative for x>0, it is strictly increasing for x>0.Also, as f(0)<0 it must have exactly 1 real root.Hence, 1/4 is the only real root of the equation.I believe this proof is not rigorous enough.More rigorous proofs (that can find all possible complex roots ) would be appreciated.