Question

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

• A. 3
• B. 12
• C. 9
• D. 4

Answer 1

Answer: (B)

We have,
$x^{3}+\frac{1}{4} x^{2}-\frac{1}{16} x+\frac{1}{144}=0$
The equation whose roots are $k$ times the roots of the given equation, is
$\left(\frac{x}{k}\right)^{3}+\frac{1}{4}\left(\frac{x}{k}\right)^{2}-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0$
$\Rightarrow x^{3}+\frac{k}{4} x^{2}-\frac{k^{2}}{16} x+\frac{k^{3}}{144}=0$
Since, the coefficients of the equation are integer.
$\therefore k=12 $