Question

The value of $mϵR$ for which both of the roots of equation $x^2-6mx+9m^2-2m+2=0$ exceed 3 is $(\frac{a}{b},\infty )$ Find $a-b$

Answer 1

Let $f\left(x\right)=0$ are greater than $3$, we can take $D\ge 0$, $af\left(3\right)>0$ and $-\frac{b}{20}>3$ ... (1)
Consider D $\ge$ 0 : ${-6m}^2-4.1({9m}^2-2m+2)\ge 0$
$\Rightarrow 8m-8\ge 0\therefore m\ge 1ormϵ(1,\infty )$...............(1)
Consider $af\left(3\right)>0:1$.$(9-18m+9m^2-2m+2)>0$
$\Rightarrow 9m^2-20m+11>0\Rightarrow (9m-11)(m-1)>0$
$\Rightarrow (m-\frac{11}{9})(m-1)>0$
$\Rightarrow mϵ(-\infty ,1)\cup (\frac{11}{9},\infty )$ ...............(2)
Consider $-\frac{b}{2a}>3:$
$\frac{6m}{2}>3\Rightarrow m>1\Rightarrow mϵ(1,\infty )$ ..........(3)
Hence, the values of $m$ satisfying (1),(2) and (3) at the same time are $mϵ(\frac{11}{9},\infty )$
$\Rightarrow a-b=2$
Ans: 2