Question

If the roots of $10x^3-c{x}^2-54x-27=0$ are in H.P., then the value of $c$ is

Answer 1

Given : The roots of $10x^3-c{x}^2-54x-27=0$ are in H.P.

Replacing $x$ by $\frac{1}{x},$ we get
$-27x^3-54x^2-cx+10=0\Rightarrow 27x^3+54x^2+cx-10=0\cdots \left(1\right)$
Now, the roots of above equation are in A.P., let those be
$a-d,a,a+d$
Sum of roots,
$3a=-2\Rightarrow a=-\frac{2}{3}$
Putting this in equation $\left(1\right)$, we get
$\Rightarrow 27{(-\frac{2}{3})}^3+54{(-\frac{2}{3})}^2+c(-\frac{2}{3})-10=0\Rightarrow -8+24-\frac{2c}{3}-10=0\therefore c=9$