Question

Find the roots of the equation $6x^2+\sqrt{2}x-2=0$.
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• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{\sqrt{2}}{2}$,$\frac{\sqrt{2}}{3}$
• B. $\frac{\sqrt{2}}{2}$,$\frac{-\sqrt{2}}{3}$
• C. $\frac{-\sqrt{2}}{2}$,$\frac{-\sqrt{2}}{3}$
• D. $\frac{-\sqrt{2}}{2}$,$\frac{\sqrt{2}}{3}$

Answer 1

The correct option is D $\frac{-\sqrt{2}}{2}$,$\frac{\sqrt{2}}{3}$

Given equation is $6x^2+\sqrt{2}x-2=0$

$\Rightarrow$ $6x^2+3\sqrt{2}x-2\sqrt{2}x-2=0$

$\Rightarrow$ $3x(2x+\sqrt{2})-\sqrt{2}(2x+\sqrt{2})=0$

$\Rightarrow$ $(2x+\sqrt{2})(3x-\sqrt{2})=0$

$\Rightarrow$ $(2x+\sqrt{2})=0$ or $(3x-\sqrt{2})=0$

$\Rightarrow$ $2x=-\sqrt{2}$ or $3x=\sqrt{2}$

$\Rightarrow$ $x=\frac{-\sqrt{2}}{2}$,$\frac{\sqrt{2}}{3}$

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