Question

Solve each of the following quadratic equations:

$\sqrt{3}{x}^2-2\sqrt{2}x-2\sqrt{3}=0$ [CBSE 2011]

Answer 1

We write, $-2\sqrt{2}x=-3\sqrt{2}x+\sqrt{2}x$ as $\sqrt{3}{x}^2\times \left(-2\sqrt{3}\right)=-6x^2=\left(-3\sqrt{2}x\right)\times \left(\sqrt{2}x\right)$

$$\begin{gathered} \therefore \sqrt{3}{x}^2-2\sqrt{2}x-2\sqrt{3}=0 \\ \Rightarrow \sqrt{3}{x}^2-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0 \\ \Rightarrow \sqrt{3}x\left(x-\sqrt{6}\right)+\sqrt{2}\left(x-\sqrt{6}\right)=0 \\ \Rightarrow \left(x-\sqrt{6}\right)\left(\sqrt{3}x+\sqrt{2}\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-\sqrt{6}=0\mathrm{or}\sqrt{3}x+\sqrt{2}=0 \\ \Rightarrow x=\sqrt{6}\mathrm{or}x=-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{6}}{3} \end{gathered}$$
Hence, the roots of the given equation are $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$.