Finding range of $m$ in $x^2+mx+6$. Find the range of values of $m$ in the quadratic equation $x^2+mx+6=0$ such that both the roots of the equation $\alpha,\beta<1$.
My attempt - it is given that
$\alpha<1$ and $\beta<1$
$\rightarrow \alpha+\beta<2$
But $\alpha+\beta=-m$
Thus $m>-2$.
But this solution doesn't involve the coefficient term i.e. $6$.
Any solution to the above question is appreciated.
 A: You made a logical fallacy as old as time.
What you have proven is:

If $\alpha, \beta<1$, then $m>-2$

What you HAVE NOT PROVEN is:

If $m>-2$, then $\alpha, \beta < 1$.

You can easily see that the second statement is false, since if $m=0>-2$, the equation has no real solutions.
A: This is a step-by-step description of how I would do it. There might be more clever ways out there, and there might be unanticipated problems with this approach, but it is the first thing I would try:


*

*Write an explicit formula for $\alpha$ and $\beta$ (it should use both the constant coefficient $6$ and the linear coefficient $m$)

*Decide which of $\alpha$ and $\beta$ is the biggest

*Set that root to be less than $1$

*Solve

A: As you see solving for $x$, one can start from the system $$\begin {cases} y=mx \\ \\ y=-x^2-6 \end {cases}$$ studying the intersections of a generic line passing through the origin ($m$ is just its slope) with a (fixed) parabola.
After checking the lines of tangency with $m=\pm \sqrt {24}$ and the line passing through the point $(1,-7)$, it is easy to deduce that both the roots are less than $1$ for every (and only) $m \ge \sqrt {24}$ .
A: 
Find the range of values of $m$ in the quadratic equation $x^2+mx+6=0$ such that both the roots of the equation $\alpha,\beta<1$.

Our graph has its wings pointing upwards. When both roots are less than one, both wings of the function's graph transect the OX axis to the left of the point $x=1$. 
In order to be sure that both transection points are to the left of $x=1$, we need to know:


*

*There exist two transection points: hence, $D>0$
(If $D=0$, there will only be one transection point, meaning only one root; if $D<0$, there will exist no transection points: no roots).  

*If we draw a vertical line from $x=1$, it will strike the graph above the OX axis: hence $f(1)>0$
(Indeed, if it strikes the graph below the OX axis, that would mean that the point $x=1$ lies between the two roots)   

*But what if it will strike the left wing of the graph, not the right one? What if the whole shebang lies to the right of $x=1$? In order to exclude this possibility, let's say that the lowest point of the graph should also lie to the left of $x=1$. Hence, $x_0<1$. The formula for $x_0$ is $x_0=\frac{-B}{2A}$


Then I would unite these three conditions in a system and solve it. 

A: Just use the pq-formula:
$$x_{1/2} = -\frac{m}{2} \pm \sqrt{\frac{m^2}{4}-6}$$
from that we ca deduce:
1) $m^2 \geq 24$, because the radiant under the root can't be negative
2)
$$-\frac{m}{2} \pm \sqrt{\frac{m^2}{4}-6} < 1$$
just solve that and you have your range
