Question

Factories the quadratic equation

Answer 1

Solution...

${3x}^{2} - 2 \sqrt{6} + 2$
${3x}^{2} - \sqrt{6} x - \sqrt{6} x + 2$
$\sqrt{3} x( \sqrt{3} x - \sqrt{2}) - \sqrt{2}( \sqrt{3} x - \sqrt{2} )$
$( \sqrt{3} x - \sqrt{2} )( \sqrt{3} x - 2)$
Therefore,
$x = \sqrt{2} \div \sqrt{3} \: or \: x = \sqrt{2} \div \sqrt{3}$
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Answer 2

Question :-

Factories it

→ 3x² -2√6x +2 = 0

Answer :-

$\implies \: \boxed{ \sf{ x \: = \frac{ \sqrt{2} }{ \sqrt{3} } }}$

Explanation :-

Method (1)

Given Quadratic equation is ↓

$\rightarrow \sf{ 3 {x}^{2} - 2 \sqrt{6} x + 2 = 0} \\ \\ \rightarrow \sf{3 {x}^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0} \\ \\ \rightarrow \sf{\sqrt{3}x ( \sqrt{3} x - \sqrt{2} ) - \sqrt{2} ( \sqrt{3} x - \sqrt{2} ) = 0} \\ \\ \rightarrow \sf{( \sqrt{3} x - \sqrt{2} )( \sqrt{3} x - \sqrt{2} ) = 0 }\\ \\ \rightarrow \sf{{( \sqrt{3}x - \sqrt{2} ) }^{2} = 0} \\ \\ \rightarrow \sf{\sqrt{3} x - \sqrt{2} = 0} \\ \\ \rightarrow \boxed{\sf{x \: = \sqrt{ \frac{2}{3} } }}$

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For verification :-

$\sf{put \: x \: = \frac{ \sqrt{2} }{ \sqrt{3} } \: in \: equation} \\ \\ \rightarrow \: 3 { \bigg( \frac{ \sqrt{2} }{ \sqrt{3} } \bigg) }^{2} - 2 \sqrt{6} \frac{ \sqrt{2} }{ \sqrt{3} } + 2 = 0 \\ \\ \rightarrow \: 3 \times \frac{2}{3} - 2 \sqrt{2} \sqrt{2} + 2 = 0 \\ \\ \rightarrow \: 2 - 4 + 2 = 0 \\ \\ \rightarrow \: 0 = 0 \\ \\ \tt hence \: verified$

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Method (2)

We have ,

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

We can write it,

$\rightarrow \sf{ {( \sqrt{3}x) }^{2} - 2 \sqrt{2} \sqrt{3} x + {( \sqrt{2}) }^{2} = 0} \\ \\ \because \sf{ {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2} } \\ \\ \rightarrow \sf{ {( \sqrt{3}x - \sqrt{2} )}^{2} = 0} \\ \\ \rightarrow \boxed{\sf{x \: = \frac{ \sqrt{2} }{ \sqrt{3} } }}$

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