Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$ 
Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$.

Discriminant$=m^2-10m+9\ge0\implies m\in(-\infty,1]\cup[9,\infty)$
And putting $m=0$, we get both roots less than $2$. Does that mean we can eliminate all the values of $m\le1?$ Why?
Putting $m=9,$ we get double root at $3$. Does that mean all values of $m\ge9$ can be accepted?
 A: 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: $m \ge 9$.
A: We have, two cases.
Case $1$: One root is smaller than $2$ and the other is greater than $2$.
$\implies 2$ lies between the roots.
This gives us a condition that: $f(2)<0$
Case $2$: Both roots are greater than $2$
This gives us the condition that: $f(2)>0$
Also, $D\ge0$ and $\frac{-b}{2a}>2$
[From graph, Vertex: $(\frac{-b}{2a}, \frac{-D}{4a})$ ]
After solving:
Case $1$ gives $m>10$, and Case $2$ gives $m \in [9,10)$
Taking union of both cases, we get:
$$\boxed{m\in [9, \infty) }$$
A: If $(f(x)=Ax^2+Bx+C=0$ has at least one root $>2$, them
There are two cases: (1) one root $>2$, (2) both roots $>2$
For $f(x)=x^2-(m-3)x+m=0$
Case 1: $B^2\ge 4AC \implies m\le 1~or~ m\ge 9$ and $f(2)<0 \implies m>10$
the overlap of the two conditions gives $n\in(10,\infty)$.
Case2: $B^2>4AC:m\le 1 ~or~ m\ge 9$ and $x_0=-B/(2A) >2 \implies (m-3)/2>2 \implies m>7$ and $f(2)>0 \implies 10-m>0 \implies m<10.$ The overlap of the three intervals is $m \in [9,10)$.
So the union of the two cases gives $m\ge 9.$
A: The quadratic polynomial $ \ x^2  −  (m−3)x  +  m  \ \ $ represents an "upward-opening" parabola, for which the "vertex form" is $ \ \left(x \  −  \ \frac{ m−3}{2} \right)^2 \ + \ \left( m \ - \ \frac{ [m−3]^2}{4} \right) \ \ . $  It may also be noted that $ \ f(1) \ = \ 4 \ \ , $ independent of the value of $ \ m \ \ . $  So there can be no zeroes of the polynomial greater than $ \ 2 \ $ at the very least for when the vertex is at $ \ x \le 1 \ \Rightarrow \ \frac{ m−3}{2} \ \le \ 1 \ \Rightarrow \ m \ \le \ 5 \ \ . $
The vertex is tangent to or "below" the $ \ x-$axis for
$$ m \ - \ \frac{ [m−3]^2}{4} \  \le \  0 \ \ \Rightarrow \ \ 0  \ \le  \ -4m \ + \ m^2 \ -6m \ + \ 9 \ \ = \ \ (m - 1)·(m - 9)  $$ $$ \Rightarrow \  \ m \ \le \ 1 \ \ , \ \ m \ \ge \ 9  $$
(as also found by other respondents).  We may thus discard the first inequality.  As for the second, $ \ m \ = \ 9 \ $ gives us the polynomial $ \ \left(x \  −  \ \frac{ 9−3}{2} \right)^2 \ = \ (x \  −  \ 3 )^2 \ \ , $ which has a "double zero" at $ \ x = 3 \ \ $ and is therefore an admissible result.  All larger values of $ \ m \ $ correspond to parabolas with vertices at $ \ x \ \ge \ 3 \ $ and $ \ y \ < \ 0 \ \ , $ so they always have an $ \ x-$intercept "to the right" of $ \ x = 3 \ \ . $  Hence, all of the quadratic polynomials $ \ x^2  −  (m−3)x  +  m  \ \ $ have at least one real zero greater than $ \ 2 \ $ for $ \ \mathbf{m \ \ge \ 9} \ \ . $
[In fact, we may note that for $ \ m = 10 \ \ , $ the polynomial becomes $ \ x^2 - 7x + 10 \ = \ (x - 2)·(x - 5) \ \ , $ so both zeroes are greater than $ \ 2 \ $ for $ \ 9 \ \le \ m \ < \ 10 \ \ . \ $ ]
