Quadratics and roots Consider the equation (E): $$x^2 - (m+1)x+m+4=0$$ where $m$ is a real parameter 
determine $m$ so that $2$ is a root of (E) and calculate the other root. 
This is the question. 
What I did was basically this:
Let the sum of root 1 and root 2 be $S$ and their product $P$
Let $x_2 = a ; x_1=2$(given)


*

*$S=m+1$ 
$m+1=2+a$ 
$m-a=1$

*$P=m+4$
$m+4=2a$
$m-2a=-4$
then these 2 equations form a system whose answers would be $m=6$ and $a=5$.
Is it possible to determine $m$ so that $x^2−(m+1)x+m+4<0$ for all $x \in  \mathbb{R}$?

If $2$ is a root of $x^2-(m+1)x+(m+4)=0$ then
\begin{align}
&2=\frac{(m+1)\pm\sqrt[\;2]{(m+1)^2-4\cdot (m+4)}}{2}
\\
\Leftrightarrow&
4=(m+1)\pm \sqrt[\;2]{m^2-2m-15}
\\
\Leftrightarrow&
-m+3 =\pm \sqrt[\;2]{m^2-2m-15}
\\
\Leftrightarrow&
(-m+3)^2 =\left(\pm \sqrt[\;2]{m^2-2m-15}\right)^2
\\
\Leftrightarrow&
(-m+3)^2 =\left|m^2-2m-15\right|
\end{align}
Case 1: If $m^2-2m-15=(m+3)(m-5)> 0$ then we have 
\begin{align}
(-m+3)^2 =+(m^2-2m-15),& \hspace{1cm} m<-3 \mbox{ or } m>5\\
m^2-6m+9 = m^2-2m-15,  & \hspace{1cm} m<-3 \mbox{ or } m>5\\
-4m = -24,  & \hspace{1cm} m<-3 \mbox{ or } m>5\\
 m = 6,  & \hspace{1cm} m<-3 \mbox{ or } m>5\\
\end{align}
Case 2: If $m^2-2m-15=(m+3)(m-5)< 0$ then we have 
\begin{align}
(-m+3)^2 =-(m^2-2m-15),&\hspace{1cm} -3<m<5\\
m^2-6m+9 =-m^2+2m+15,&\hspace{1cm} -3<m<5\\
2m^2-8m-6 =0,&\hspace{1cm} -3<m<5\\
m^2-4m-3 =0,&\hspace{1cm} -3<m<5\\
m=\frac{4\pm \sqrt{4^2-4\cdot 1\cdot (-3)}}{2\cdot 1},&\hspace{1cm} -3<m<5\\
m=2\pm \sqrt{7},&\hspace{1cm} -3<m<5\\
\end{align}
In this case there is no integer solution.
Case 3: If $m^2-2m-15=(m+3)(m-5)=0$ then we have 
\begin{align}
(-m+3)^2 =0,& \hspace{1cm} m=-3 \mbox{ or } m=5\\
    m=3,& \hspace{1cm} m=-3 \mbox{ or } m=5\\
\end{align}
But it's impossible.
