Question

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

Answer 1

Answer: 4

$\because 3=2\left(\frac{c}{a}-\frac{b}{a}\right)$
$\Rightarrow 3=2\left(4{\alpha }^2+5\alpha \right)\Rightarrow 8{\alpha }^2+10\alpha -3=0$
$\Rightarrow 8{\alpha }^2+12\alpha -2\alpha -3=0$
$\Rightarrow (2\alpha +3)(4\alpha -1)=0$
$\therefore \alpha =\frac{1}{4}\Rightarrow \beta =4\alpha =1$
$\therefore S=\frac{\beta }{1-\alpha }=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}\Rightarrow 3S=4$