Question

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

are $\frac{-1}{\sqrt{2}},2\sqrt{2}$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$\Rightarrow$ Here, $a=1,b=-\frac{3}{\sqrt{2}},c=-2$

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

${\left(\frac{-b}{2a}\right)}^2={\left(\frac{-\left(\frac{-3}{\sqrt{2}}\right)}{2\left(1\right)}\right)}^2=\frac{9}{8}$

Adding $\frac{9}{8}$ on both sides,

$\Rightarrow$ $x^2-\frac{3}{\sqrt{2}}x+\frac{9}{8}=2+\frac{9}{8}$

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

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

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

$\Rightarrow$ $x=2\sqrt{2}$ and $x=-\frac{1}{\sqrt{2}}$