Question

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

Answer 1

Given equation : $10x^3-c{x}^2-54x-27=0$

let roots be $p,q,r$

$\therefore p+q+r=\frac{c}{10}$ .............. $\left(1\right)$

$pq+qr+pr=\frac{-54}{10}$ ............... $\left(2\right)$

$pqr=\frac{27}{10}$ ........... $\left(3\right)$

$\because$ Roots are in $H.P.$

$\therefore \frac{1}{p},\frac{1}{q},\frac{1}{r}$ in $A.P.$

$\frac{2}{q}=\frac{1}{p}+\frac{1}{r}$

or $2pr=q(p+r)$ ......... $\left(4\right)$

$2pr=pq+qr\Rightarrow 2pr=\frac{-54}{10}-pr$

or, $3pr=\frac{-54}{10}\Rightarrow pr=\frac{-18}{10}$

But $pqr=\frac{27}{10}\Rightarrow -\frac{18}{10}q=\frac{27}{10}\Rightarrow q=\frac{-3}{2}$

from $\left(4\right)$, $2\left(\frac{-18}{10}\right)=\frac{-3}{2}(p+r)\Rightarrow p+r=\frac{24}{10}$

$=\frac{24}{10}-\frac{3}{2}=\frac{c}{10}\Rightarrow c=9$ Ans