Question

The roots of the equation $3x^2-2\sqrt{6}x+2=0$ are
(a) $\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}$
(b) $\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}$
(c) $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer 1

(b) $\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}$

$$\begin{gathered} \text{Given:} \\ 3x^2-2\sqrt{6}x+2=0 \\ \mathrm{We}\mathrm{know}\mathrm{that}: \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \text{=}\frac{2\sqrt{6}\pm \sqrt{{(-2\sqrt{6})}^2-4\times 3\times 2}}{2\times 3} \\ \text{=}\frac{2\sqrt{6}\pm \sqrt{24-24}}{6} \\ \text{=}\frac{2\sqrt{6}\pm 0}{6} \\ \therefore x=\frac{2\sqrt{6}}{6}=\frac{\sqrt{6}}{3}=\frac{\sqrt{3}\times \sqrt{2}}{3}=\sqrt{\frac{2}{3}} \\ \text{Therefore, the roots are}\sqrt{\frac{2}{3}}\text{and}\sqrt{\frac{2}{3}}\text{.} \end{gathered}$$