How do I the roots of a quadratic function given certain conditions?

Question

Let's say I have an equation $x^2+2(1-m)x+6m-11=0$, how would I go about finding for what values of m are both roots located in the segment -1<x<1?

Answer 1

By the quadratic formula,

$x=\dfrac{-2(1-m)\pm\sqrt{4(1-m)^2-4(6m-11)}}{2}$

$=-(1-m)\pm\sqrt{(1-m)^2-(6m-11)}$

$=-(1-m)\pm\sqrt{m^2-8m+12}$

First, by examining the determinant, we have

$m^2-8m+12\geq0$

$(m-4)^2-4\geq0$

$m\geq6\text{ or }m\leq2\ \ldots(1)$

Then, regarding the range of $x$,

$-1<x<1$

$-1<-(1-m)\pm\sqrt{m^2-8m+12}<1$

$-m<\pm\sqrt{m^2-8m+12}<2-m$

Since a positive square root is non-negative and a negative square root is non-positive, we require that

$\begin{cases} -m\leq0 \\ 2-m\geq0 \end{cases}$

Hence, $0\leq m\leq2\ \ldots(2)$.

For positive square root,

$0<+\sqrt{m^2-8m+12}<2-m$

$0\leq m^2-8m+12<(2-m)^2$

$m^2-8m+12<4-4m+m^2$

$4m>8$

$m>2\ \ldots(3)$

For negative square root,

$-m<-\sqrt{m^2-8m+12}\leq0$

$0\leq m^2-8m+12<m^2$

$-8m+12<0$

$m>\frac{3}{2}\ \ldots(4)$

Conclusion

Now, we have the compound inequalities:

$\begin{cases} m\geq6 \text{ or } m\leq2\ \ldots(1)\\ 0\leq m\leq2\ \ldots(2)\\ m>2\ \ldots(3)\\ m>\frac{3}{2}\ \ldots(4)\\ \end{cases}$

Since (2) & (3) contradict each other, there are no solutions for $m$.

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Regarding your suggestion of making $f(1),f(-1)>0$ followed by testing the determinant constraint, I was quite amazed at first glance until I gave it a second thought. There is a flaw in this method.

Substituting $m=6$, we have

$f(x)=x^2-10x+25$

$f(1)=16>0$ and $f(-1)=36>0$, whereas the root is $x=5$.

That means, if you only guarantee $f(1),f(-1)>0$, hoping that the two roots are between them, you fail to rule out the possibility that $f(1)$ and $f(-1)$ are on the same side of the roots, and hence the roots fall out of range.

Answer 2

Putting the quadratic polynomial into "vertex form" gives us $$ x^2 \ + \ 2(1 \ - \ m)·x \ + \ (6m \ - \ 11) \ \ = \ \ ( \ x \ - \ [m - 1] \ )^2 \ + \ (6m \ - \ 11) \ + \ (m \ - \ 1)^2 $$ $$ = \ \ ( \ x \ - \ [m - 1] \ )^2 \ + \ (6m \ - \ 11) \ - \ (m \ - \ 1)^2 \ \ = \ \ (x \ - \ [m - 1])^2 \ - \ (m^2 - 8m + 12) $$ $$ = \ \ (x \ - \ [m - 1])^2 \ - \ (m - 2)·(m - 6) \ \ . $$

As with Hirofumi Ryo's method using the discriminant of the polynomial, we see that there are no zeroes for $ \ (m - 2)·(m - 6) \ < \ 0 \ \Rightarrow \ 2 \ < \ m \ < \ 6 \ \ . \ $ Further, we can reject $ \ m \ \ge \ 6 \ \ , \ $ as this places the vertex of the represented parabola at $ \ x \ \ge \ 5 \ \ , \ $ so certainly at least one of the zeroes is greater than or equal to $ \ 5 \ \ . \ $

But $ \ m \ \le \ 2 \ $ must also be rejected. With $ \ m \ = \ 2 \ \ , \ $ the vertex lies at $ \ x \ = \ 1 \ \ , \ $ so while the polynomial has a "double zero" at $ \ x \ = \ 1 \ \ , \ $ this is not located in the interval $ \ (-1 \ , \ 1) \ \ . \ $ Decreasing $ \ m \ $ brings at most one zero into the specified interval, but not both. For example, at $ \ m \ = \ \frac{11}{6} \ \ , \ $ the polynomial becomes $ \ x^2 - \frac53·x \ \ , \ $ with its zeroes at $ \ x \ = \ 0 \ $ and $ \ x \ = \ \frac53 \ \ . \ $ For $ \ m \ = \ \frac32 \ \ , \ $ the polynomial is $ \ x^2 - x - 2 $ $ = \ (x + 1)·(x - 2) \ \ , \ $ with zeroes at $ \ x \ = \ -1 \ $ and $ \ x \ = \ +2 \ \ . \ $ So as the vertex "moves downward and to the left", the larger of the two zeroes remains greater than or equal to $ \ +1 \ \ , \ $ never entering the interval in question; meanwhile, the smaller zero is less than $ \ -1 \ $ for $ \ m \ < \ \frac32 \ \ . \ $ Thus we find that there is no value of $ \ m \ $ which can place both zeroes in $ \ (-1 \ , \ 1) \ \ . $

[With a little additional work, we can show that the larger zero $ \ (m \ - 1) + \sqrt{m^2-8m+12} $ $ = \ (m \ - 1) + \sqrt{(m -4)^2 - 4} \ $ asymptotically approaches $ \ +3 \ \ : $ $$ \lim_{m \ \rightarrow \ -\infty} \ (m \ - 1) + \sqrt{(m -4)^2 - 4} \ \ = \ \ (m \ - \ 1) \ - \ (m \ - \ 4) \ \ = \ \ 3 \ \ . \ ] $$