Question

On Solving the following quadratic equation by factorization, the roots are :$\sqrt{6},-\sqrt{\frac{2}{3}}$
$\sqrt{3}{x}^2-2\sqrt{2x}-2\sqrt{3}=0$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

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

$\therefore$ We can see, roots given in question are correct.