Question

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


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Answer 1

Answer:

$\mathtt\red{x = \frac{ - 2}{3} }$

Step-by-step explanation:

$\tt({3x}^{2} + 2 \sqrt{6}x + 2 = 0 \\ {3x}^{2} + \sqrt{6x} + \sqrt{6x} + 2 = 0 \\ \sqrt{3x} ( \sqrt{3x} + \sqrt{2}) + \sqrt{2}( \sqrt{3x} + \sqrt{2} ) \\ (\sqrt{3x} + \sqrt{2} )( \sqrt{3x} + \sqrt{2} ) \\ {(\sqrt{3 x} + \sqrt{2} )}^{2} \\ ( \sqrt{3x} + \sqrt{2} ) = 0 \\ ( \sqrt{3x} + \sqrt{2} ) = 0 \\ \sqrt{3x} = - \sqrt{2} \\ 3x = - 2 \\ \pink{ x = \frac{ - 2}{3} }$