<s>[INST] $x^2 - (m - 2)x + 6 = 0$ : Possible values of $m$. 
One of the roots of the equation $x^2 - (m - 2)x + 6 = 0$ lies between
$(0, 2)$ and the other root lies between $(3, 4)$. It is known that
$m$ is an integer. What is the sum of all the possible values of $m$?

Let $a$ be the root that lies between $(0, 2)$ and let $b$ be the root that lies between $(3, 4)$. I can write it mathematically as
$0 \lt a \lt2$ and $3 \lt b \lt4$ and also we can say that $3 \lt a+b \lt6$.
$a+b=m-2$ and this means that $3 \lt m-2 \lt6 \Rightarrow 5 \lt m \lt 8$. So the possible integer values of $m$ should be $6$ and $7$ and the possible sum should be $13$.
But as per the solution this is not the correct answer. What have I done wrong? Please help !!!
Thanks in advance !!!
 [/INST] No real root is possible for integral $m$ in (0,2) and (3,4)
Let $f(x)=x^2-(m-2)x+6$. $f(0)=6>0$, then for a root in (0,2) we should have $f(2))<0 \implies 14-2m<0 \implies m>7$.
Next for a root in (3,4), $f(3)=9-3(m-2)+6=21-3m<0$, next we demand $f(4)>0 \implies 16-4(m-2)+6>0\implies m<7.5$. Hence no integral value of $m$ is possible.
For $m=7$ the roots are 2 and 3. For $7 <m<7.5$, roots lie in (0,2) and (3,4).Hence, no real root is possible for integral m in (0,2) and (3,4).
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