<s>[INST] 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?
 [/INST] 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 \ \ . \ $ ]
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