Not as bad as it looks

In the given equation, if $x$ is a solution, then so is $-x$ . Therefore, $S = 0$ .

Ok my solution $\sqrt{12-\frac{3}{x^{2}}}+\sqrt{4x^2-\frac{3}{x^2}}=4x^2$ Let $x^{2}=a> 0$ $\Rightarrow \sqrt{12-\frac{3}{a}}+\sqrt{4a-\frac{3}{a}}=4a$ $\Leftrightarrow \sqrt{4a-\frac{3}{a}}=4a-\sqrt{12-\frac{3}{a}}$ $\Rightarrow 4a-\frac{3}{a}=16a^2-8a\sqrt{12-\frac{3}{a}}+12-\frac{3}{a}$ $\Leftrightarrow 4a^{2}-2a\sqrt{12-\frac{3}{a}}-a+3=0$ $\Leftrightarrow 4a^{2}-a-\sqrt{48a^{2}-12a}+3=0$ $\Leftrightarrow 12a(4a-1)-12\sqrt{12a(4a-1)}+36=0$ Let $\sqrt{12a(4a-1)}=y\geq 0$

$\Rightarrow y^2-12y+36=0 \Rightarrow y=6$ $\Rightarrow 48a^2-12a=36 \Rightarrow a=1$ $a=-\frac{3}{4}$ Because of $a> 0$ So $a= 1$ $\Rightarrow x=\pm 1$ $S=1-1=0$