Find $m\in\mathbb{R}$ such that inequality $x^2-(m+2)x+5-m<0$ is valid $\forall~x\in(1;3)$.
How I think of it:
Set $x^2-(m+2)x+5-m=0$
The parabola will be oriented upwards. In order to satisfy the hypothesis we need the following conditions:
$x_1\leq1,~x_2\geq3$
Write coefficients:
$ \begin{cases} a=1 \\ b=-(m+2) \\ c=5-m \end{cases} $
Calculate discriminant:
$\Delta=b^2-4ac=m^2+8m-16$
Put the condition $\Delta>0$ so the equation will have 2 distinct real roots:
$m^2+8m-16>0$
Set $m^2+8m-16=0$ and find the roots:
$m_1=-4(\sqrt{2}+1),~m_2=4(\sqrt{2}-1)$
$m\in(-\infty;m_1)\cup(m_2;\infty)\Rightarrow m\in\big(-\infty;-4(\sqrt{2}+1)\big)\cup\big(4(\sqrt{2}-1);\infty\big)$
Now find the roots of the initial equation:
$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{m+2\pm\sqrt{\Delta}}{2}$
Substitute in the conditions:
$\frac{m+2-\sqrt{\Delta}}{2}\leq1\Rightarrow m\leq\sqrt{\Delta}$
$\frac{m+2+\sqrt{\Delta}}{2}\geq3\Rightarrow -m+4\leq\sqrt{\Delta}$
Substitute $\Delta$ and solve the system of inequalities:
$ \begin{cases} m\leq\sqrt{m^2+8m-16} \\ -m+4\leq\sqrt{m^2+8m-16} \end{cases} $
We can square both sides unless the left hand side is negative. Example : $-2<1$ but $(-2)^2<1^2$ is false. How can I proceed? One idea would be to make a table of signs for $m$ and $-m+4$ and take in consideration all the possible cases, then take the reunion at the end. Is there an easier way to solve these equations with square roots? Also, am I overcomplicating this problem? Thanks for the help
You have $(\forall x\in(1,3)):f(x)<0$ if and only if $f(1),f(3)\leqslant0$. Since $f(1)=4-2m$ and $f(3)=8-4m$, you have $f(1),f(3)\leqslant0$ if and only if $4-2m\leqslant0$, which is equivalent to $m\geqslant2$.