Quadratics and roots

Question

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)

1. $S=m+1$

   $m+1=2+a$

   $m-a=1$

2. $P=m+4$

   $m+4=2a$

   $m-2a=-4$

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 $x \in \mathbb{R}$?

Answer 1

Divide the given polynomial by $x-2$.

This yields the quotient $x-m+1$ and the remainder $-m+6$.

Then $m=6$ and $x=5.$

Answer 2

Your solution is correct, but you should use more standard variable names for your roots.

Use the Vieta's formulas. $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$

Comparison with your equation yields: $$x_1+x_2=m+1$$ $$x_1x_2=m+4$$ $$\text{subtraction}\implies x_1x_2-(x_1+x_2)=4-1$$

Now, plug in $x_1=2$.

Answer 3

plugging $2$ into the given equation we get $$4-2(m+1)+m+4=0$$ or $$m=6$$ and for $m=6$ we get $$x^2-7x+10=0$$

Answer 4

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.