Question

The given quadratic equations have real roots and roots are Real and equal, $\sqrt{\frac{3}{2}}$ :
$2x^2-2\sqrt{6x}+3=0$

• A. True
• B. False

Answer 1

The correct option is A True

$2x^2-2\sqrt{6}x+3=0$

$\Rightarrow$ $2x^2-\sqrt{6}x-\sqrt{6}x+3=0$

$\Rightarrow$ $\sqrt{2}x(\sqrt{2}x-\sqrt{3})-\sqrt{3}(\sqrt{2}x-\sqrt{3})=0$ [ Since, $\sqrt{6}=\sqrt{3}\times \sqrt{2}$ ]

$\Rightarrow$ $(\sqrt{2}x-\sqrt{3})(\sqrt{2}x-\sqrt{3})=0$

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

$\therefore$ $x=\frac{\sqrt{3}}{\sqrt{2}}$ and $x=\frac{\sqrt{3}}{\sqrt{2}}$

$\therefore$ We can see, roots given in question are correct.