(A) A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Let the natural number be x

\[\begin{aligned} & \text { ATQ, } x+12=\frac{160}{x} \\ & x^{2}+12 x=160 \\ & x^{2}+12 x-160=0 \\ & (x+20)(x-8)=0 \\ & x \neq-20, x=8 \end{aligned} \]

⇒ Required natural number is $\displaystyle 8$

\[x^{2}+12 x-k=0 \]

Let the roots be \(\displaystyle \alpha, 3 \alpha\)

\[\alpha+3 \alpha=-12 \Rightarrow \alpha=-3 \]

\(\displaystyle \alpha \cdot 3 \alpha=-\mathrm{k} \quad \Rightarrow 3 \alpha^{2}=-\mathrm{k}\)

\[\Rightarrow k=-27 \]