For the existence of the roots, we need $m\le 1$ or $m \ge 9$.
The roots are
$$\frac{(m-3) \pm \sqrt{(m-3)^2-4m}}{2}$$
We want the larger root to be more than $2$,
$$\frac{m-3+\sqrt{(m-3)^2-4m}}{2}>2$$
$$\sqrt{m^2-10m+9} > 7-m$$
Clearly, any $m \ge 9$ would satisties the inequality as the RHS is negative and LHS is positive.
If $m \le 1$, $$m^2-10m+9 > m^2-14m+49$$
$$m > 10$$
and we find that the intersection is empty.
Summary: $m \ge 9$.
This answer is nothing more than a consolidation of all of the responses.
To solve: $ax^2 + bx + c \geq 0.$
Since it is presumed that $a \neq 0$, the first thing to do is to break the analysis into two cases: $a < 0$ or $a > 0$. Then, similar to the manner of deriving the quadratic equation, in each case, you must complete the square.
In both cases, I will use the observation that
$x^2 + \left(\frac{b}{a}\right)x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}.$
I will also use the observation that when $r,s \geq 0$, you have that
$(r \geq s) \iff (r^2 \geq s^2).$
$\underline{\text{Case 1:} ~a > 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \geq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \geq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \geq \frac{b^2 - 4ac}{4a^2}.\tag1$$
Within Case 1: there are therefore two possibilities:
Case 1A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (1) above, then $x$ must satisfy:
$\left|x + \frac{b}{2a}\right| \geq \frac{\sqrt{b^2 - 4ac}}{2a}$.
This is satisfied when
$x \leq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{2a}$
or
$x \geq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{2a}$.
Case 1B:
$b^2 - 4ac < 0$.
Examining inequality (1) above, since case 1B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (1) will be satisfied for any value of $x$.
$\underline{\text{Case 2:} ~a < 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \leq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \leq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \leq \frac{b^2 - 4ac}{4a^2}.\tag2$$
Case 2A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (2) above, then $x$ must satisfy:
$\left|x + \frac{b}{2a}\right| \leq \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
This is satisfied when
$x \geq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{|2a|}$
and
$x \leq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
Case 2B:
$b^2 - 4ac < 0$.
Examining inequality (2) above, since case 2B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (2) will never be satisfied for any value of $x$.
Best Answer