Question

Find the roots of the following quadratic equations:

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

Answer 1

Step 1:

Find the two roots of the quadratic equation:

Compare given quadratic equation with $a{x}^2+bx+c=0$,

we get, $a=\sqrt{3},b=-2\sqrt{2},c=-2\sqrt{3}$

Now, using quadratic formula:

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4\times a\times c}}{2a} \\ \Rightarrow x=\frac{-\left(-2\sqrt{2}\right)\pm \sqrt{{\left(-2\sqrt{2}\right)}^2-4\times \sqrt{3}\times -2\sqrt{3}}}{2\times \sqrt{3}} \\ \Rightarrow x=\frac{2\sqrt{2}\pm \sqrt{8+8\times 3}}{2\times \sqrt{3}} \\ \Rightarrow x=\frac{2\sqrt{2}\pm \sqrt{32}}{2\times \sqrt{3}} \\ \Rightarrow x=\frac{2\sqrt{2}\pm 4\sqrt{2}}{2\times \sqrt{3}} \\ \Rightarrow x=\frac{2\sqrt{2}+4\sqrt{2}}{2\times \sqrt{3}},\frac{2\sqrt{2}-4\sqrt{2}}{2\times \sqrt{3}} \\ \Rightarrow x=\frac{6\sqrt{2}}{2\sqrt{3}},\frac{-2\sqrt{2}}{2\sqrt{3}} \\ \Rightarrow x=\frac{3\sqrt{2}}{\sqrt{3}},\frac{-\sqrt{2}}{\sqrt{3}} \\ \Rightarrow x=\sqrt{3}\times \sqrt{2},\frac{-\sqrt{2}}{\sqrt{3}}\left[\because 3=\sqrt{3}\times \sqrt{3}\right] \\ \therefore x=\sqrt{6},-\sqrt{\frac{2}{3}} \end{gathered}$$

Final Answer:

The roots of the quadratic equation are $\sqrt{6},-\sqrt{\frac{2}{3}}$.