Question

Let $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ where $\beta =4\alpha (\alpha >0)$. If $3a=2(c-b)$ and $S=\sum _{r=0}^{\infty }\beta \left({\alpha }^r\right)$, then find the value of $18S$.

Answer 1

Answer: 24

$5\alpha =\frac{-b}{a}\text{ and }4{\alpha }^2=\frac{c}{a}$
$\therefore b=-5a\alpha \text{ and }c=4a{\alpha }^2$
$\text{ Now }3a=2(c-b)$
$3a=2\left(4a{\alpha }^2+5a\alpha \right)$
$\therefore 8{\alpha }^2+10\alpha -3=0$
$\alpha =\frac{1}{4}\text{ or }\alpha =\frac{-3}{2}\text{ (rejected) }$
$\text{ hence }\beta =1$
$\text{ Now }S=\beta (1+\alpha +\dots \dots .+\infty )$
$S=\frac{\beta }{1-\alpha }=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
$18S=24$