Question

Solve each of the following quadratic equations:

$3x^2-2\sqrt{6}x+2=0$ [CBSE 2010, 12]

Answer 1

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

$$\begin{gathered} \therefore 3x^2-2\sqrt{6}x+2=0 \\ \Rightarrow 3x^2-\sqrt{6}x-\sqrt{6}x+2=0 \\ \Rightarrow \sqrt{3}x\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}x-\sqrt{2}\right)=0 \\ \Rightarrow \left(\sqrt{3}x-\sqrt{2}\right)\left(\sqrt{3}x-\sqrt{2}\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(\sqrt{3}x-\sqrt{2}\right)}^2=0 \\ \Rightarrow \sqrt{3}x-\sqrt{2}=0 \\ \Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3} \end{gathered}$$
Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.