Find the roots of the following quadratic equ | vedclass

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(D) To solve the quadratic equation $16 x^{2}-24 x-1=0$ by the method of completing the square,we first rewrite the equation as $16 x^{2}-24 x = 1$.Di

Vedclass · Accepted Answer

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

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(D) To solve the quadratic equation $16 x^{2}-24 x-1=0$ by the method of completing the square,we first rewrite the equation as $16 x^{2}-24 x = 1$.Divide the entire equation by the coefficient of $x^{2}$,which is $16$:$x^{2} - \frac{24}{16} x = \frac{1}{16}$$x^{2} - \frac{3}{2} x = \frac{1}{16}$Now,add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{3}{2}$,so half of it is $-\frac{3}{4}$,and its square is $\frac{9}{16}$:$x^{2} - \frac{3}{2} x + \frac{9}{16} = \frac{1}{16} + \frac{9}{16}$This simplifies to:$(x - \frac{3}{4})^{2} = \frac{10}{16}$Taking the square root of both sides:$x - \frac{3}{4} = \pm \sqrt{\frac{10}{16}}$$x - \frac{3}{4} = \pm \frac{\sqrt{10}}{4}$Solving for $x$:$x = \frac{3}{4} \pm \frac{\sqrt{10}}{4}$$x = \frac{3 \pm \sqrt{10}}{4}$Thus,the roots are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$.

Find the roots of the following quadratic equation by the method of completing the square: $16 x^{2}-24 x-1=0$

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