I am trying to find the range of values for $m$ such that the equation $n^4-2n^2m-6n^2+2m^2-2m=0$ will have 4 real and distinct roots. My approach so far has been:
Let $u=n^2$ (I feel that this substitution shouldn't make a difference as I'm not trying to find the specific roots of $n$)
$ \implies u^2-2um-6u+2m^2-2m$ = 0
$ \implies u^2+(-2m-6)u+(2m^2-2m)$ = 0
Using the quadratic discriminant $b^2-4ac$, I obtain:
$(-2m-6)^2 - (4)(1)(2m^2-2m) > 0 $
$ \implies -4m^2+32m+36>0$
$ \implies m^2-8m-9<0 $
$ \implies (m-9)(m+1)<0 $
$ \implies -1<m<9 $
All of this seemed fine to me, however the answer appears to be $-1<m<0$ AND $1<m<9$. In effect, my inequality is correct except between 0 and 1 inclusive. The only way I found this out was by graphing the original equation. How can I derive the correct inequalities ALGEBRAICALLY?
It should be $$2m+6>0,$$ $$2m^2-2m>0$$ and $$(m+3)^2-(2m^2-2m)>0.$$
Let $n^2=x$.
Hence, we need to find all values of $m$ for which the equation $$x^2-2(m+3)x+2m^2-2m=0$$ has two positive roots.
The quadratic equation $ax^2+bx+c=0$ has positive roots $x_1$ and $x_2$ iff $$x_1+x_2=-\frac{b}{a}>0,$$ $$x_1x_2=\frac{c}{a}>0$$ and $$\Delta=b^2-4ac>0.$$