Question

For what values of $m\in R$, both roots of the equation $x^2-6mx+9m^2-2m+2=0$ exceed $3$?

• A. $[-7,-2]$
• B. $(\frac{11}{9},\infty )$
• C. $[1,\infty ]$
• D. $[0,\infty ]$

Answer 1

The correct option is B $(\frac{11}{9},\infty )$

Given quadratic equation $x^2-6mx+9m^2-2m+2=0$
Condition for both roots greater than a number $k$ is $D\ge 0,-\frac{b}{2a}>k,af\left(k\right)>0$
(i) Consider $D\ge 0$${(-6m)}^2-4(9m^2-2m+2)\ge 0$
$\Rightarrow 8m-8\ge 0$
$\Rightarrow m\ge 1$
$\Rightarrow m\in [1,\infty )$ .....(1)
(ii) Consider $f\left(3\right)>0$
$1.(9-18m+19m^2-2m+2)>0$
$\Rightarrow 9m^2-20m+11>0$
$\Rightarrow (9m-11)(m-1)>0$
$\therefore (m-\frac{11}{9})(m-1)>0$
$\Rightarrow m\in (-\infty ,1)\cup (\frac{11}{9},\infty ).....\left(2\right)$
(iii) Consider $-\frac{b}{2a}>3\Rightarrow \frac{6m}{2}>3$
$m>1$
$\Rightarrow m\in (1,\infty )$ .....(3)
Hence, from $\left(1\right),\left(2\right),\left(3\right)$ $m\in (\frac{11}{9},\infty )$
Hence, option 'B' is correct.