Question

Solve the equation by formula method.
$\sqrt{3}{x}^2+2\sqrt{2}x-2\sqrt{3}=0$

Answer 1

Given equation is,

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

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

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

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

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

$⟹x=\frac{-2\sqrt{2}\pm \sqrt{32}}{2\sqrt{3}}$

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

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

$⟹x=\frac{-\sqrt{2}\pm 2\sqrt{2}}{\sqrt{3}}$

$⟹x=\frac{-\sqrt{2}+2\sqrt{2}}{\sqrt{3}}$ or $x=\frac{-\sqrt{2}-2\sqrt{2}}{\sqrt{3}}$

$⟹x=\frac{\sqrt{2}}{\sqrt{3}}$ or $x=\frac{-3\sqrt{2}}{\sqrt{3}}$

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