Obtain the roots of the following equation us | vedclass

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Question

(C) Given equation: $16x^{2} - 24x - 1 = 0$.Divide the entire equation by $16$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} - \frac{24}{16}x

Vedclass · Accepted Answer

$\frac{3-\sqrt{10}}{4}, \frac{3+\sqrt{10}}{4}$

Vedclass · Answer

(C) Given equation: $16x^{2} - 24x - 1 = 0$.Divide the entire equation by $16$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} - \frac{24}{16}x - \frac{1}{16} = 0 \implies x^{2} - \frac{3}{2}x - \frac{1}{16} = 0$.Move the constant term to the right side:$x^{2} - \frac{3}{2}x = \frac{1}{16}$.Add the square of half the coefficient of $x$ (which is $\frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$,and its square is $\frac{9}{16}$) to both sides:$x^{2} - \frac{3}{2}x + \left(\frac{3}{4}ight)^{2} = \frac{1}{16} + \frac{9}{16}$.Rewrite the left side as a perfect square:$\left(x - \frac{3}{4}ight)^{2} = \frac{10}{16}$.Take the square root of both sides:$x - \frac{3}{4} = \pm \sqrt{\frac{10}{16}} = \pm \frac{\sqrt{10}}{4}$.Solve for $x$:$x = \frac{3}{4} \pm \frac{\sqrt{10}}{4} = \frac{3 \pm \sqrt{10}}{4}$.Thus,the roots are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$.

Obtain the roots of the following equation using the method of 'completing the square': $16x^{2} - 24x - 1 = 0$.

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