Question

three x square -2 root 6 +2 = 0 फैक्टराइजेशन मेथड​

Answer 1

$\huge\sf\red{\underline{Answer}}$

$3 {x}^{2} - 2 \sqrt{6}x + 2 = 0 \\ = > 3 {x}^{2} - \sqrt{2 \times 3}x - \sqrt{2 \times 3}x + 2 = 0 \\ = > \sqrt{3}x( \sqrt{3}x - \sqrt{2} ) - \sqrt{2} ( \sqrt{3}x - \sqrt{2}) = 0 \\ = > ( \sqrt{3}x - \sqrt{2} )( \sqrt{3} x - \sqrt{2} ) = 0$

So,

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

Answer 2

$\huge{ \underline{ \bold{ᴀɴsᴡᴇʀ....{ \heartsuit}}}}$

$$\begin{lgathered}3 {x}^{2} - 2 \sqrt{6}x + 2 = 0 \\ = > 3 {x}^{2} - \sqrt{2 \times 3}x - \sqrt{2 \times 3}x + 2 = 0 \\ = > \sqrt{3}x( \sqrt{3}x - \sqrt{2} ) - \sqrt{2} ( \sqrt{3}x - \sqrt{2}) = 0 \\ = > ( \sqrt{3}x - \sqrt{2} )( \sqrt{3} x - \sqrt{2} ) = 0\end{lgathered}$$

So,

$$\begin{lgathered}\sqrt{3} x - \sqrt{2} = 0 \\ = > \sqrt{3} x = \sqrt{2} \\ = > x = \frac{ \sqrt{2} }{ \sqrt{3} } \\ = > x = \sqrt{ \frac{2}{3} }\end{lgathered}$$