Question

On solving the following quadratic equation by factorization, the roots are $\frac{-1}{\sqrt{2}},2\sqrt{2}$:
$\sqrt{2}{x}^2-3x-2\sqrt{2}=0$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$\Rightarrow$ $\sqrt{2}x(x-2\sqrt{2})+1(x-2\sqrt{2})=0$

$\Rightarrow$ $(x-2\sqrt{2})(\sqrt{2}x+1)=0$

$\Rightarrow$ $x-2\sqrt{2}=0$ and $\sqrt{2}x+1=0$

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

$\Rightarrow$ $\frac{-1}{\sqrt{2}},2\sqrt{2}$ are roots of the given equation $\sqrt{2}{x}^2-3x-2\sqrt{2}=0$ is true.