Question

If the roots of $x^{3}-k x^{2}+14 x-8=0$ are in geometric progression, then $k$ is equal to

• A. -3
• B. 7
• C. 4
• D. 0

Answer 1

Answer: (B)

Given equation is
$x^{3}-k x^{2}+14 x-8=0$
According to the question,
$\frac{a}{r}, a$ and $a r$ is the roots of equation.
Then, $\frac{a}{r} \cdot a \cdot a r=\frac{D}{A}$
$\Rightarrow \frac{a}{r} \cdot a \cdot a r=8 \Rightarrow a^{3}=8$
$\Rightarrow a=2$
Since, $a=2$ is a root of given equation.
$\therefore (2)^{3}-k(2)^{2}+14(2)-8=0$
$\Rightarrow 8-4 k+28-8=0$
$ \Rightarrow k=7$