Finding range of $m$ in $x^2+mx+6$.

Question

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.

Answer 1

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.

Answer 2

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:

1. Write an explicit formula for $\alpha$ and $\beta$ (it should use both the constant coefficient $6$ and the linear coefficient $m$)
2. Decide which of $\alpha$ and $\beta$ is the biggest
3. Set that root to be less than $1$
4. Solve

Answer 3

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:

1. 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).
2. 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)
3. 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.

Answer 4

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) $m \ge \sqrt {24}$ .

Answer 5

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