Question

root 3x square - 2 root 2x - 2 root 3

Answer 1

$\sqrt{3}x {}^{2} - 2 \sqrt{2}x - 2 \sqrt{3} = 0 \\ =>\sqrt{3}x {}^{2} - 3 \sqrt{2}x + \sqrt{2}x - 2 \sqrt{3} = 0 \\ => \sqrt{3}x(x - \sqrt{6}) + \sqrt{2}(x - \sqrt{6}) = 0 \\ =>(x - \sqrt{6})( \sqrt{3}x + \sqrt{2}) = 0$
$= > x = \sqrt{6} \: \: and \: \: - \frac{ \sqrt{2} }{ \sqrt{3} }.$


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