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Question

(C) Given equation: $6x^2 + 11x + 3 = 0$Divide the entire equation by $6$ to make the coefficient of $x^2$ equal to $1$:$x^2 + \frac{11}{6}x + \frac{3

Vedclass · Accepted Answer

$-\frac{3}{2}, -\frac{1}{3}$

Vedclass · Answer

(C) Given equation: $6x^2 + 11x + 3 = 0$Divide the entire equation by $6$ to make the coefficient of $x^2$ equal to $1$:$x^2 + \frac{11}{6}x + \frac{3}{6} = 0 \implies x^2 + \frac{11}{6}x + \frac{1}{2} = 0$Shift the constant term to the right side:$x^2 + \frac{11}{6}x = -\frac{1}{2}$Add the square of half the coefficient of $x$ (i.e.,$(\frac{1}{2} \cdot \frac{11}{6})^2 = (\frac{11}{12})^2 = \frac{121}{144}$) to both sides:$x^2 + \frac{11}{6}x + \frac{121}{144} = -\frac{1}{2} + \frac{121}{144}$$(x + \frac{11}{12})^2 = -\frac{72}{144} + \frac{121}{144} = \frac{49}{144}$Taking the square root on both sides:$x + \frac{11}{12} = \pm \frac{7}{12}$Case $1$: $x = \frac{7}{12} - \frac{11}{12} = -\frac{4}{12} = -\frac{1}{3}$Case $2$: $x = -\frac{7}{12} - \frac{11}{12} = -\frac{18}{12} = -\frac{3}{2}$Thus,the roots are $-\frac{3}{2}$ and $-\frac{1}{3}$.

Obtain the roots of the following equation using the method of 'completing the square': $6x^2 + 11x + 3 = 0$

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