Finding $m$ so that the roots of $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$ 
Find $m$ so that the roots of the equation $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$.

Having $x_1, x_2 \in (1, 5)$, we get $x_1 + x_2 \in (2, 10)$ and $x_1x_2 \in (1, 25)$. By solving these two inequations I got $m \in \left( \sqrt3, \frac{11}{3} \right)$. However, this result is wrong. By substituting $m$ with $2$, for example, the given conditions are not reached.
So, please, explain what I did wrong and how to get the right solution.
Thank you in advance!
 A: There is a neat method I learnt to solve such questions. Let $f(x)=x^2-(3m-1)x+m^2-2$. We observe that the function is an "open-up" parabola, that is, it is going to have a minima at some point and goes to infinity in both directions. Now, we need that the vertex of the parabola (that is, the point of minima) must lie in the interval $(1,5)$. The abscissa of the vertex can be found easily by differentiation, or by direct formula $x=-b/2a$ (for a general parabola $ax^2+bx+c=0$).
Also, when both the roots lie in the interval $(1,5)$, clearly $f(1)>0$ and $f(5)>0$. This gives two more inequalities. Finally, the discriminant of the quadratic must be greater than zero, giving us the final inequality. Note that geometrically, the vertex is $\left(\dfrac{-b}{2a}, \dfrac{-D}{4a}\right)$, where $D$ is the discriminant, and in our case $a>0$, so the vertex must lie below x-axis, that is why we have the condition $D>0$. Similar reasoning would apply for $a<0$. So the inequalities we have are:
$$ 1<\dfrac{3m-1}{2}<5 \\
   m^2-2-(3m-1)+1>0 \\
   m^2-2-5(3m-1)+25>0 \\
   (3m-1)^2-4(m^2-2)>0  $$
Other answers have elaborated enough that no $m$ satisfies this system of inequalities.
The advantage of this method is that it works for other quadratics too, and it gives the complete interval where $m$ can possibly lie. Also, it is less messy.
A: To get the right solution, Solve the equation with respect to $m$:
$$x_{1,2}=\frac{1}{2} \left(3 m-1\pm\sqrt{5 m^2-6 m+9}\right)$$
You can see for yourself that both $x_1$ and $x_2$ are strictly increasing functions of $m$. The proof is not hard.

Now take a look at the above picture to see that there is no such $m$.
A: Notice that $\Delta = 5m^2 -6m +9 > 0, \forall m \in \mathbb{R}.$ Hence, 
$x = \dfrac{3m-1 \pm \sqrt{5m^2 -6m +9}}{2} \in \mathbb{R}.$ 
$(1):~~~\dfrac{3m-1 - \sqrt{5m^2 -6m +9}}{2} > 1 \implies (3m-3)^2 > 5m^2 -6m +9 \implies 4m(m -3) > 0 \implies  m < 0 ~~\text{or}~~ m > 3.$ But, only $ m > 3$ satisfies the condition (we squared the inequality!).
$(2):~~~ \dfrac{3m-1 + \sqrt{5m^2 -6m +9}}{2} < 5 \implies 5m^2 -6m +9 <(-3m+11)^2 \implies m^2 -15m +28 > 0 \implies m < \dfrac{15-\sqrt{113}}{2} \approx 2.185 ~~\text{or}~~ m > \dfrac{15+\sqrt{113}}{2} \approx 12.185.$ However, only $m < \dfrac{15-\sqrt{113}}{2}$ is solution of the inequality $(2)$.
Thus, combine the results of $(1)$ and $(2)$, we notice that there is no $m$ such that $x \in (1,5)$.
A: Using your approach, you have to solve the system:
$$
\begin{cases}
3m-1\ge 2\\
3m-1\le 10\\
m^2-2\ge 1\\
m^2-2 \le 25
\end{cases}
$$
$$
\begin{cases}
m\ge 1\\
m\le\frac{11}{3}\\
m\le -\sqrt{3} \lor m\ge\sqrt{3}\\
m \le -\sqrt{23}
 \lor m \ge \sqrt{23}
\end{cases}
$$
It is not difficult to show that this has no solution (can you do?) .
From your post it is not possible to see where is your mistake.
