<s>[INST] Find the value of $a$ such that at least one root of the equation $f(x)=x^2 - (a-3)x + a =0$ is greater than $2$. As the title says.
To solve this problem I took two cases and solved them separately:
1. when the $x$ coordinate of the vertex is greater than $2$ and $f(2)>0$; 
2. when $f(2)<0$.
However I wasn't able to find the answer as the two conditions have no common solution (obviously they won't have because in one case $f(2)>0$ and in the other one $f(2)<0$. 
The inequalities which I got after solving for the two cases separately were: 
1. $a<10$ and $a<7$ 
2. $10 < a$ 

And help would be appreciated. :)
 [/INST] $x^2-(a-3)+a=0$
$\therefore \alpha, \beta=\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{(a-3)\pm \sqrt{(a-3)^2-4a}}{2}=\frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}$
$$\therefore\large \frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}>2$$
$$\therefore(a-3)\pm \sqrt{(a-1)(a-9)}>4$$
$$\therefore \pm \sqrt{(a-1)(a-9)} > 4-a+3$$
$$\therefore (a-1)(a-9) > (7-a)^2$$
$$\therefore (a-1)(a-9) \geq 0$$
For this to be true either both brackets on LHS should be positive OR both negative.
$\therefore a \geq9$ OR $a \leq1$
EDIT: substitute $a=9$ and $a=1$ in the original equation and we see that $a=9$ satisfies the constraint while $a=1$ does not, so answer is $a \geq9$
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