Question

Solve the following quadratic equation by factorization :
$3x^2-2\sqrt{6x}+2=0$

The roots are $\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3}$

• A. True
• B. False

Answer 1

The correct option is B False

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

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

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

$\Rightarrow$ $3x^2-\left(\sqrt{2}\right)\left(\sqrt{3}\right)x-\left(\sqrt{2}\right)\left(\sqrt{3}\right)x+2=0$

$\Rightarrow$ $\sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})=0$

$\Rightarrow$ $(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})=0$

$\Rightarrow$ $\sqrt{3}x-\sqrt{2}=0$ and $\sqrt{3}x-\sqrt{2}=0$

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

'$\therefore$ Here we can see, roots given in question are incorrect.