Question

solve the following quadratic equation by factorisation method​

Answer 1

here is your answer..

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Answer 2

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

We will factorize

Using splitting the middle term method

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

$\longmapsto \sf {3x}^{2} - \sqrt{2 \times 3}x - \sqrt{2 \times 3}x + 2 = 0$

$\longmapsto \sf {3x}^{2} - ( \sqrt{2} )( \sqrt{3})x - ( \sqrt{2} )( \sqrt{3})x + 2 = 0$

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

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

$\longmapsto \sf \sqrt{3}x - \sqrt{2} = 0$

$\longmapsto \sf \sqrt{3}x = \sqrt{2}$

$\longmapsto \sf x = \dfrac{ \sqrt{2} }{ \sqrt{3} }$

$\longmapsto\boxed{ \sf x = \sqrt{ \dfrac{2}{3}}}$