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.
$\therefore k=12$