Question

Find the roots of 6x²-√2x-2=0 b factorisation

Answer 1

Step-by-step explanation:

$6 {x}^{2} - \sqrt{2} x - 2 = 0 \\ = > 6 {x}^{2} - 3 \sqrt{2} x + 2 \sqrt{2} x - 2 = 0 \\ = > 3x(2x - \sqrt{2} ) + \sqrt{2} (2x - \sqrt{2}) = 0 \\ = > (2x - \sqrt{2})(3x + \sqrt{2} ) = 0 \\ now \: either \: (2x - \sqrt{2} ) = 0 \\ = > \fbox{x = \frac{ \sqrt{2} }{2} = \frac{1}{ \sqrt{2} } } \\ or \: (3x + \sqrt{2} ) = 0 \\ = > \fbox{ x = \frac{ - \sqrt{2} }{3} }$

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