Question

$3 {x}^{3} - 2 \sqrt{6x} + 2 = 0$
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Answer 1

Answer :

$\sqrt{ \frac{2}{3} } \: \: \: \: \sqrt{ \frac{2}{3} }$

Step-by-step explanation:

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

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

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

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

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

$= > x = \frac{ \sqrt{2} }{ \sqrt{3} } \: \: \: \: \: \: \: \: x = \frac{ \sqrt{2} }{ \sqrt{3} }$