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(B) Let the given equation be $x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{144}=0$ $(i)$.If the roots of the new equation are $k$ times the roots of e

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$12$

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(B) Let the given equation be $x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{144}=0$ $(i)$.If the roots of the new equation are $k$ times the roots of equation $(i)$,we replace $x$ with $\frac{x}{k}$ in equation $(i)$.$\left(\frac{x}{k}\right)^3+\frac{1}{4}\left(\frac{x}{k}\right)^2-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0$Multiplying the entire equation by $k^3$ to clear the denominators:$x^3+\frac{k}{4} x^2-\frac{k^2}{16} x+\frac{k^3}{144}=0$For the coefficients to be integers,$k$ must be such that $4$ divides $k$,$16$ divides $k^2$,and $144$ divides $k^3$.Checking the options:If $k=12$,then $\frac{k}{4} = \frac{12}{4} = 3$,$\frac{k^2}{16} = \frac{144}{16} = 9$,and $\frac{k^3}{144} = \frac{1728}{144} = 12$.Since all coefficients $(1, 3, -9, 12)$ are integers,$k=12$ is a possible value.

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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