For which values of $m$ we get: exist $\alpha$ and $\beta$ roots of this equation, so that: $-1< \alpha < 1 < \beta < 2$?

Question

given this equation:
$$(m+1)x^2 -2(m-1)x + m=0$$

For which values of $m$ we get: exist $\alpha$ and $\beta$ roots of this equation, so that: $-1< \alpha < 1 < \beta < 2$?

Answer 1

If you plug in $x=1$ into $f(x)=(m+1)x^2-2(m-1)x+m$ you get $f(1)=3$. Since you have $-1\lt\alpha\lt1$ and $f(\alpha)=0$, you can infer that $f(-1)\lt0$. Similarly $f(2)\lt0$. The function changes sign on those intervals because you have exactly one root in each interval. You should get a system of two inequalities that would constrain the value of $m$.

Answer 2

Let $f(x)=(m+1)x^2-2(m-1)x+m$.

We need to solve the following system. $$\{m|m>-1,f(-1)>0,f(1)<0,f(2)>0\}\cup\{m|m<-1,f(-1)<0,f(1)>0,f(2)<0\}.$$ Since $f(1)=3>0$, we work with the right set only, which gives $m<-8$.