Question

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

Answer 1

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= √3x² + √6 = 9

Squaring both sides

= (√3x² + √6 )² = 9²

= 3x² + 6 + 2√18x² - 81 = 0

= 3x² + 6x√2 - 75 = 0

= 3 ( x² + 2x√2 - 25 ) = 0

= x² + 2x√2 - 25 = 0

= Discriminant = 2√2² - 4*-25

= Discriminant = 8 + 100 = 108

= √Discriminant = √108 = 6√3

So roots = - b + √D / 2a, - b - √D / 2a

= 2√2 + 6√3 / 2 , 2√2 - 6√3 / 2

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