Show your love for radicals!

$\begin{aligned} \sqrt{x-3}+\sqrt{2x-2} & = 2 \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ x-3 + 2\sqrt{(x-3)(2x-2)}+2x-2 & = 4 \quad \quad \small \color{#3D99F6}{\text{Rearranging}} \\ 3x - 9 & = -2\sqrt{(x-3)(2x-2)} \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ 9(x - 3)^2 & = 4(x-3)(2x-2) \quad \quad \small \color{#3D99F6}{\text{Rearranging}} \\ (x-3)(9x - 27 - 8x + 8) & = 0 \\ (x-3)(x - 19) & = 0 \\ \implies x & = \begin{cases} \color{#3D99F6}{3} & \implies \sqrt{\color{#3D99F6}{3}-3}+\sqrt{2(\color{#3D99F6}{3})-2} \color{#3D99F6}{= 2} & \color{#3D99F6}{\text{Accepted}} \\ \color{#D61F06}{19} & \implies \sqrt{\color{#D61F06}{19}-3}+\sqrt{2(\color{#D61F06}{19})-2} = 10 \color{#D61F06}{\ne 2} & \color{#D61F06}{\text{Rejected}} \end{cases} \end{aligned}$

The sum of solutions is $\boxed{3}$ .

We have, $\displaystyle \sqrt{x-3}+ \sqrt{2x-2} = 2$

$\implies \displaystyle \sqrt{x-3}+ \sqrt{2} \times \sqrt{x-1} = 2$

Let $u = \sqrt{x-1} \implies u^{2} = x-1 \implies u^{2} - 2 = x-3$

Hence, our equation becomes:

$\sqrt{u^{2}-2} = 2 - \sqrt{2}u$

Squaring both sides, we have:

$u^{2}-2=4-4 \sqrt{2}u + 2u^{2} \implies u^{2} -4 \sqrt{2}u + 6 = 0 \implies u= \sqrt{2}, 3 \sqrt{2}$

Since $u= \sqrt{x-1}$ , we have $x=3$ and $x=19$ . But substituting these values in the primitive equation, we find that only $x=3$ is a solution but $x=19$ isn't. So we find that the only legitimate solution to the equation is $3$ .