Question

9x^-2✓6x+2=0 solve it
$9x {}^{2} + \sqrt{6x } + 2 = 0$

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Answer 1

given=>

$\starblack{9x^2-2√6-2}$

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solution=>

Step-by-step explanation:

$= > 9x {}^{2} - 2 \sqrt{6}x - 2 = 0 \\ \\ = > 9x {}^{2} - 3 \sqrt{6} x + \sqrt{6} - 2 = 0 \\ \\ = > 3 \sqrt{3} x( \sqrt{3} x - \sqrt{2)} \\ \: \: \: + \sqrt{2} ( \sqrt{3} x - \sqrt{2} ) = 0 \\ \\ = > (3 \sqrt{3}x + \sqrt{2} ) ( \sqrt{3}x - \sqrt{2} ) = 0$

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now compair x value

$= > 3 \sqrt{3} x + \sqrt{2} = 0 \\ = > 3 \sqrt{3} x = - \sqrt{2} \\ = > x = \frac{ - \sqrt{2} }{3 \sqrt{3} } \\ and = > \sqrt{3}x - \sqrt{2 } = 0 \\ \sqrt{3} x = \sqrt{2} \\ = > x = \frac{ \sqrt{2} }{ \sqrt{3} }$

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I hope it help