Question

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

Answer 1

$5 \alpha=\frac{-b}{a} \text { and } 4 \alpha^2=\frac{c}{a} $
$\therefore b =-5 a \alpha \text { and } c =4 a \alpha^2 $
$\text { Now } 3 a =2( c - b )$
$3 a=2\left(4 a \alpha^2+5 a \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 $
$\text { Now } S=\beta(1+\alpha+\ldots \ldots .+\infty)$
$S =\frac{\beta}{1-\alpha}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
$18 S =24$