Question

Solve each of the following equations by using the method of completing the square:

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

Answer 1

$$\begin{gathered} \sqrt{2}{x}^2-3x-2\sqrt{2}=0 \\ \Rightarrow 2x^2-3\sqrt{2}x-4=0\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{2}\right) \\ \Rightarrow 2x^2-3\sqrt{2}x=4 \\ \Rightarrow {\left(\sqrt{2}x\right)}^2-2\times \sqrt{2}x\times \frac{3}{2}+{\left(\frac{3}{2}\right)}^2=4+{\left(\frac{3}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{3}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(\sqrt{2}x-\frac{3}{2}\right)}^2=4+\frac{9}{4}=\frac{25}{4}={\left(\frac{5}{2}\right)}^2 \\ \Rightarrow \sqrt{2}x-\frac{3}{2}=\pm \frac{5}{2}\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow \sqrt{2}x-\frac{3}{2}=\frac{5}{2}\mathrm{or}\sqrt{2}x-\frac{3}{2}=-\frac{5}{2} \\ \Rightarrow \sqrt{2}x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\mathrm{or}\sqrt{2}x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1 \end{gathered}$$
$\Rightarrow x=\frac{4}{\sqrt{2}}=2\sqrt{2}\mathrm{or}x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$
Hence, $2\sqrt{2}$ and $-\frac{\sqrt{2}}{2}$ are the roots of the given equation.