Question:
$48 x^{2}-13 x-1=0$
Solution:
Given:
$48 x^{2}-13 x-1=0$
$\Rightarrow 48 x^{2}-(16 x-3 x)-1=0$
$\Rightarrow 48 x^{2}-16 x+3 x-1=0$
$\Rightarrow 16 x(3 x-1)+1(3 x-1)=0$
$\Rightarrow(16 x+1)(3 x-1)=0$
$\Rightarrow 16 x+1=0$ or $3 x-1=0$
$\Rightarrow x=\frac{-1}{16}$ or $x=\frac{1}{3}$
Hence, the roots of the equation are $\frac{-1}{16}$ and $\frac{1}{3}$.
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