Question

find the positive of
$\sqrt{3x { }^{2} } + \sqrt{6 = 9}$

Answer 1

Hi mate ,

$\sqrt{3 {x}^{2} } + \sqrt{6} = 9 \\ = > \sqrt{3 {x}^{2} } = 9 - \sqrt{6} \\ = > 3 {x}^{2} = 81 + 6 - 9 \sqrt{6} \\ = > {x}^{2} = \frac{87 - 9 \sqrt{6} }{3} \\ = > {x}^{2} = 29 - 3 \sqrt{6} \\ = > x = \sqrt{29 - 3 \sqrt{6} }$
it is the positive value of x

can you help me .

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