Question

If one root of equation $(l-m)x^2+lx+1=0$ be double of the other and if $l$ be real, then $m\le \frac{a}{b}$ where $a$ and $b$ are integers in the simplest form. Find $Min(a+b)$.

Answer 1

$Thesum$ $of$ $roots$ $is$ $\frac{-l}{(l-m)}=3r,$

$Theproduct$ $of$ $roots$ $is$ $\frac{1}{(l-m)}=2r^2$ $where$ $r$ $is$ $one$ $of$ $the$ $roots.$

$So,$ $\frac{1}{2(l-m)}=\frac{l^2}{9{(l-m)}^2}$

$9(l-m)=2l^2$

$2l^2-9l+9m=0$

$For$ $l$ $to$ $real,$ $the$ $discriminant$ $is$ $non-negative.$

$81\ge 8\times 9m$

$m\le \frac{9}{8}.$