Question

For the given equation $\sqrt{2}{x}^2-3x+\sqrt{2}=0$, the value of roots are

• A. $2,\frac{1}{\sqrt{2}}$
• B. $\sqrt{2},\frac{1}{\sqrt{2}}$
• C. $\sqrt{2},1$
• D. $-\sqrt{2},\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\sqrt{2},\frac{1}{\sqrt{2}}$

$\sqrt{2}{x}^2-3x+\sqrt{2}=0$
Dividing L.H.S and R.H.S by $\sqrt{2}$.
$\Rightarrow x^2-\frac{3}{\sqrt{2}}x+1=0$
$\Rightarrow x^2-\frac{1}{\sqrt{2}}x-\frac{2}{\sqrt{2}}x+1=0$
$\Rightarrow x^2-\frac{1}{\sqrt{2}}x-\sqrt{2}x+1=0$
$\Rightarrow x(x-\frac{1}{\sqrt{2}})-\sqrt{2}(x-\frac{1}{\sqrt{2}})=0$
$\Rightarrow (x-\frac{1}{\sqrt{2}})(x-\sqrt{2})=0$
$\therefore \text{Roots are}\frac{1}{\sqrt{2}}\text{and}\sqrt{2}$