Question

The roots of the following quadratic equation Real and equal.

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

$\therefore$ We can see roots are real and equal.