An Interesting equation!

From the equation

$\begin{aligned} \left \{ \lfloor x \rfloor + \frac x{\sqrt 3} \right \} - \left \{\frac {\sqrt 3}x \right \} & = 0 \\ \left \{ \frac x{\sqrt 3} \right \} - \left \{\frac {\sqrt 3}x \right \} & = 0 \\ \implies \left \{\frac x{\sqrt 3} \right \} & = \left \{\frac {\sqrt 3}x \right \} \end{aligned}$

A trivial solution is $x = \sqrt 3$ as $1 < \sqrt 3 < 2$ . To find other solution, let us consider for $1 \le x \le 2$ , $\dfrac x{\sqrt 3}$ increases from $\dfrac 1{\sqrt 3}$ to $\dfrac 2{\sqrt 3}$ and $\dfrac x{\sqrt 3} = 1$ , when $x = \sqrt 3$ ; and $\dfrac {\sqrt 3}x$ decreases from $\dfrac {\sqrt 3}1$ to $\dfrac {\sqrt 3}2$ and $\dfrac {\sqrt 3}x = 1$ , again when $x = \sqrt 3$ .

Therefore we have $\begin{cases} \left \{ \dfrac x{\sqrt 3} \right \} = \begin{cases} \dfrac x{\sqrt 3} & \text{for } 1 \le x < \sqrt 3 \\ \dfrac x{\sqrt 3} - 1 & \text{for } \sqrt 3 \le x \le 2 \end{cases} \\ \left \{ \dfrac {\sqrt 3}x \right \} = \begin{cases} \dfrac {\sqrt 3}x - 1 & \text{for } 1 \le x < \sqrt 3 \\ \dfrac {\sqrt 3}x & \text{for } \sqrt 3 \le x \le 2 \end{cases} \end{cases}$

The following figure show the two functions {\left {\frac x{\sqrt 3} \right } in red and {\left{ \frac {\sqrt 3}x \right }) in blue.

The solution for $1 \le x < \sqrt 3$ is given be:

$\begin{aligned} \frac x{\sqrt 3} & = \frac {\sqrt 3}x - 1 \\ x^2 + \sqrt 3 x - 3 & = 0 \\ x & = \frac {-\sqrt 3 + \sqrt{15}}2 \end{aligned}$

The solution for $\sqrt 3 \le x \le 2$ is the trivial solution $x = \sqrt 3$ . Therefore, twice the sum of two roots is $\sqrt 3 + \sqrt{15}$ and $a+b = 3+5 = \boxed 8$ .

{ $\lfloor x\rfloor +\frac{x}{\sqrt 3}$ }-{ $\frac{\sqrt 3}{x}$ } $=0$

$\implies$ { $\frac{x}{\sqrt 3}$ }={ $\frac{\sqrt 3}{x}$ }

One obvious solution is $x=\sqrt 3$ , when both sides go zero. For the other solution(s), let

$\frac{x}{\sqrt 3}=α,\frac{\sqrt 3}{x}=1+α$ , where $0<α<1$ (the other possibility, viz. $\frac{\sqrt 3}{x}=α,\frac{x}{\sqrt 3}=1+α$ is forbidden, since in that case the condition $x<\sqrt 3$ will not be valid)

Solving for $α$ , we get

$α=\dfrac {-1\pm \sqrt 5}{2}$

Since $0<α<1,α=\dfrac {\sqrt 5-1}{2}$

So the two roots of the equation are

$x=\sqrt 3,x=\sqrt 3\times \dfrac {\sqrt 5-1}{2}=\dfrac {\sqrt {15}-\sqrt 3}{2}$ ,

and twice their sum is $\sqrt 3+\sqrt {15}=\sqrt 3+\sqrt {3\times 5}$

Hence the required answer is $3+5=\boxed 8$ .

It is the solution..