Question

Find the roots of the following quadratic equation:

$\sqrt{2}{x}^2-4\sqrt{3}x+4\sqrt{2}=0$

Answer 1

Find the two roots of the quadratic equation.

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

$\begin{array}{rcl}a& =& \sqrt{2}\\ b& =& -4\sqrt{3}\\ c& =& 4\sqrt{2}\end{array}$

Now, using quadratic formula:

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-\left(4\times a\times c\right)}}{2a} \\ \Rightarrow x=\frac{-\left(-4\sqrt{3}\right)\pm \sqrt{{\left(-4\sqrt{3}\right)}^2-\left(4\times \sqrt{2}\times 4\sqrt{2}\right)}}{2\times \sqrt{2}} \\ \Rightarrow x=\frac{4\sqrt{3}\pm \sqrt{48-32}}{2\times \sqrt{2}} \\ \Rightarrow x=\frac{4\sqrt{3}\pm \sqrt{16}}{2\times \sqrt{2}} \\ \Rightarrow x=\frac{4\sqrt{3}\pm 4}{2\times \sqrt{2}} \\ \Rightarrow x=\frac{4\left(\sqrt{3}\pm 1\right)}{2\times \sqrt{2}} \\ \Rightarrow x=\sqrt{2}\left(\sqrt{3}\pm 1\right) \\ \therefore x=\sqrt{6}\pm \sqrt{2} \end{gathered}$$

Hence, the roots of the quadratic equation are $\left(\sqrt{6}+\sqrt{2}\right)\mathrm{and}\left(\sqrt{6}-\sqrt{2}\right)$.