Question

Find all values of $a$ for which both roots of the equation $x^2-6ax+2-2a+9a^2=0$ are greater than $3$.

Answer 1

$x^2-6ax+2-2a+9a^2=f\left(x\right)=0$

Both roots are greater than 3.

$\frac{6a}{2}>3$

$3a-3>0$

$a>1$

$aϵ(1,\infty )$

$b>0$

$(6a{)}^2-4(2-2a+9a^2)>0$

$9a^2-2-2a-9a^2>0$

$a-1>0$

$aϵ(1,\infty )$

$f\left(3\right)>0$

$3^2-18a+2-2a+9a^2>0$

$9a^2-20a+11>0$

$9a^2-9a-11a+11>0$

$(9a-11)(a-1)>0$

$aϵ(-\infty ,1)\cup (\frac{11}{9}.\infty )$

So, $aϵ(\frac{11}{9},\infty )$