Find the roots of 6x^{2} - sqrt{2}x - 2 = 0 b | vedclass

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(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.We need two numbers whose product is $6 	imes (-2) = -12$

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$\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{3}$

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(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.We need two numbers whose product is $6 	imes (-2) = -12$ and whose sum is $-\sqrt{2}$.These numbers are $-3\sqrt{2}$ and $2\sqrt{2}$.$6x^{2} - 3\sqrt{2}x + 2\sqrt{2}x - 2 = 0$$3x(2x - \sqrt{2}) + \sqrt{2}(2x - \sqrt{2}) = 0$$(3x + \sqrt{2})(2x - \sqrt{2}) = 0$Setting each factor to zero:$3x + \sqrt{2} = 0 \implies x = -\frac{\sqrt{2}}{3}$$2x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{2}$Thus,the roots are $-\frac{\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{2}$.

Find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$ by the factorisation of the corresponding quadratic polynomial.

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