Question

Find the root of the following quadratic equation (if they exist) by the method of completing the square.
$\sqrt{2}{x}^2-3x-2\sqrt{2}=0$

Answer 1

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

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

$x^2-\frac{2.3}{2\sqrt{2}}x+\frac{9}{8}=2+\frac{9}{8}$

$(x-\frac{3}{2\sqrt{2}}{)}^2=\frac{25}{8}$

$x-\frac{3}{2\sqrt{2}}=\pm \frac{5}{2\sqrt{2}}$

$x=\frac{8}{2\sqrt{2}}$ or $\frac{1}{\sqrt{2}}$