I have this "symmetric" quadratic equation: \begin{align*} a(1-x)^2+b(1-x)x+c x^2 = 0 \end{align*} and I am trying to impose conditions on $a,b,c$ such that there is at least one root $0<x_* <1$. In that case, writing $y=1-x$ the equation reads as: \begin{align*} a y^2+b(1-y)y+c (1-y)^2 = 0 \end{align*} (hence the "symmetry"). I expect something like $ac<0$. Any ideas/reference?
\Cheers!
To be sure there is at least a root between $0$ and $1$ ,By IVT, it is sufficient to have $f (0)f (1)<0$ or $$ac <0$$
You might have to deal with this one using the good ol' quadratic formula. First we need to expand your quadratic: $$a(1−x)^2+b(1−x)x+cx^2=0$$ $$a(x^2-2x+1)+b(x-x^2)+cx^2=0$$ $$ax^2-2ax+a+bx-bx^2+cx^2=0$$ $$(a-b+c)x^2+(b-2a)x+a=0$$ Now we can use the quadratic formula: $$x=\frac{2a-b\pm \sqrt{(b-2a)^2-4a(a-b+c)}}{2(a-b+c)}$$ You're looking for a zero between $0$ and $1$, so the relationship you're looking for is $$0<\frac{2a-b\pm \sqrt{(b-2a)^2-4a(a-b+c)}}{2(a-b+c)}<1$$ $$0<2a-b\pm \sqrt{(b-2a)^2-4a(a-b+c)}<2a-2b+2c$$ $$b-2a<\pm \sqrt{(b-2a)^2-4a(a-b+c)}<b+2c$$ $$(b-2a)^2<(b-2a)^2-4a(a-b+c)<(b+2c)^2$$ $$b^2-4ab+4a^2<b^2-4ab+4a-4a^2+4ab-4ac<b^2+4bc+4c^2$$ $$-4ab+4a^2<4a-4a^2-4ac<4bc+4c^2$$ $$a-ab^2<a-a^2-ac<bc+c^2$$ That should be it, but if I made a mistake in my algebra or if I can simplify it further, please let me know.
