Solve for $x:\ \sqrt{3} x^{2} -2\sqrt{2} x-2\sqrt{3} =0$."
33207"

Given: The equation: $\sqrt{3} x^{2} -2\sqrt{2} x-2\sqrt{3} =0$.

To do: To find the value of x.

$\sqrt{3} x^{2} -2\sqrt{2} x-2\sqrt{3} =0$

On comparing the given equation to $ax^{2} +bx+c=0$

We have, $a=\sqrt{3} ,\ b=-2\sqrt{2} ,\ c=-2\sqrt{3}$

By using the formula,

$x=\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

$\Rightarrow x=\frac{-\left( -2\sqrt{2}\right) \pm \sqrt{\left( -2\sqrt{2}\right)^{2} -4\times \sqrt{3} \times -2\sqrt{3}}}{2\times \sqrt{3}}$

$\Rightarrow x=\frac{2\sqrt{2} \pm \sqrt{8+24}}{2\sqrt{3}}$

$\Rightarrow x=\frac{2\sqrt{2} \pm 4\sqrt{2}}{2\sqrt{3}}$

$\Rightarrow x=\frac{2(\sqrt{2} \pm 2\sqrt{2)}}{2\sqrt{3}}$

$\Rightarrow x=\frac{(\sqrt{2} \pm 2\sqrt{2)}}{\sqrt{3}}$

If $x=\frac{(\sqrt{2} +2\sqrt{2)}}{\sqrt{3}}$

$\Rightarrow x=\frac{3\sqrt{2}}{\sqrt{3}} =\sqrt{3} \times \sqrt{2}$

$\Rightarrow x=\sqrt{6}$

If $x=\frac{\sqrt{2} -2\sqrt{2}}{\sqrt{3}}$

$\Rightarrow x=-\frac{\sqrt{2}}{\sqrt{3}}$

Thus, $x=\sqrt{6}$ or $-\sqrt{\frac{2}{3}}$