If one root of the equation $(a-b)x^2+ax+1=0$ be double the other and if $a\in R,$ then the greatest value of $8b$, is

Let the roots be $\alpha$ and $2 \alpha .$. Then,

$\begin{array}{l}\alpha +2\alpha =-\frac{a}{a-b}\text{ and }2{\alpha }^2=\frac{1}{a-b}\\ \Rightarrow \alpha =-\frac{a}{3(a-b)}\text{ and }{\alpha }^2=\frac{1}{2(a-b)}\\ \Rightarrow \frac{a^2}{9(a-b{)}^2}=\frac{1}{2(a-b)}\\ \Rightarrow 2a^2=9a-9b\\ \Rightarrow 2a^2-9a+9b=0\\ \Rightarrow 81-72b>0 [\because a\in R]\\ \Rightarrow b\le 9/8\end{array}$

Hence, the greatest value of $8b$ is 9 .