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If the coefficients of the equation whose roots are $k$ times the roots of the equation, $x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{144}=0$ are integers, then a possible value of $k$ is
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Verified Answer
The correct answer is:
$12$
Given equation,
The equation whose roots are $\mathrm{k}$ times the roots of the eq. (i), is
$\begin{aligned}
& \left(\frac{x}{k}\right)^3+\frac{1}{4}\left(\frac{x}{k}\right)^3-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0 \\
& \Rightarrow x^3+\frac{k}{4} x^2-\frac{k^2}{16} x+\frac{k^3}{144}=0
\end{aligned}$
Also given that, coefficients of the above equation are integer. $k=12$
The equation whose roots are $\mathrm{k}$ times the roots of the eq. (i), is
$\begin{aligned}
& \left(\frac{x}{k}\right)^3+\frac{1}{4}\left(\frac{x}{k}\right)^3-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0 \\
& \Rightarrow x^3+\frac{k}{4} x^2-\frac{k^2}{16} x+\frac{k^3}{144}=0
\end{aligned}$
Also given that, coefficients of the above equation are integer. $k=12$
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