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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 6 · Set 3 · Q27
A natural number, when increased by $\displaystyle 12$, equals $\displaystyle 160$ times its reciprocal. Find the number.If one root of the quadratic equation $\displaystyle x^{2}+12 x-\mathrm{k}=0$ is thrice the other root, then find the value of k.
A natural number, when increased by $\displaystyle 12$, equals $\displaystyle 160$ times its reciprocal. Find the number.
If one root of the quadratic equation $\displaystyle x^{2}+12 x-\mathrm{k}=0$ is thrice the other root, then find the value of k.
Marking-scheme solution
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 \(\displaystyle \alpha, 3 \alpha\)
\[\alpha+3 \alpha=-12 \Rightarrow \alpha=-3
\]
\(\displaystyle \alpha \cdot 3 \alpha=-\mathrm{k} \quad \Rightarrow 3 \alpha^{2}=-\mathrm{k}\)
\[\Rightarrow k=-27
\]
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