Question

Find the root of the following quadratic equation
$3x^2-2\sqrt{6}x+2=0$

Answer 1

$3x^2-2\sqrt{6}x+2=0$

$\Rightarrow$ Here, $a=3,b=2\sqrt{6},c=2$

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

$\Rightarrow$ $x=\frac{-(-2\sqrt{6})\pm \sqrt{(-2\sqrt{6}{)}^2-4(3\left)\right(2)}}{2\left(3\right)}$

$=\frac{2\sqrt{6}\pm \sqrt{24-24}}{6}$

$=\frac{2\sqrt{6}\pm 0}{6}$

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