Solve this


$16 x^{2}=24 x+1$



Given :

$16 x^{2}=24 x+1$

$\Rightarrow 16 x^{2}-24 x-1=0$

On comparing it with $a x^{2}+b x+c=0$

$a=16, b=-24$ and $c=-1$

Discriminant $D$ is given by:

$D=\left(b^{2}-4 a c\right)$

$=(-24)^{2}-4 \times 16 \times(-1)$



Hence, the roots of the equation are real,

Roots $\alpha$ and $\beta$ are given by :

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-24)+\sqrt{640}}{2 \times 16}=\frac{24+8 \sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10})}{4}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-24)-\sqrt{640}}{2 \times 16}=\frac{24-8 \sqrt{10}}{32}=\frac{8(3-\sqrt{10})}{32}=\frac{(3-\sqrt{10})}{4}$

Thus, the roots of the equation are $\frac{(3+\sqrt{10})}{4}$ and $\frac{(3-\sqrt{10})}{4}$.


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