Mathematics - Quadratic Equation Question with Solution | TestHub

∵ Roots of $10x^3-c{x}^2-54x-27=0$ are in HP

Replacing $x$ by $\frac{1}{x}$, then we get $\frac{10}{x^3}-\frac{c}{x^2}-\frac{54}{x}-27=0$

Or $27xh3+54x^2+cx-10=0$

Now, roots of Eq (i) are in AP

Let roots $\alpha -\beta ,\alpha ,\alpha +\beta$, then $\alpha -\beta +\alpha +\alpha +\beta =-\frac{54}{27}=-2$ or $\alpha =-\frac{2}{3}$

$$\begin{array}{rlrl}& \because \alpha =-\frac{2}{3}\text{ is a root of Eq.(i) then }27{\left(-\frac{2}{3}\right)}^3+54{\left(-\frac{2}{3}\right)}^2+c\left(-\frac{2}{3}\right)-10=0& & \end{array}$$

$$or-8+24-\frac{2c}{3}-10=0\therefore c=9$$