If $x_1$ is a root of $ax^2+bx+c=0$ , $x_2$ is a root of $-ax^2+bx+c=0$ where $0<x_1<x_2$, show that the equation $ax^2+2bx+2c=0$ has a root satisfying $0<x_1<x_3<x_2$.
I have been trying this question with different approaches like using vieta's formula, plotting a rough graph of these equations(locating their roots and vertex) and manipulating the given equations to find any relation between $x_1,x_2,x_3$ but i fail to prove the above.I hope for an elegant solution to the given problem. Thanks.
Given that $$ax_1^2+bx_1+c==0~~~(1),~~ -ax_2^2+bx_2+c=0,~~(2)$$ Next $$f(x)=ax^2+2bx+2c \implies f(x_1)=ax_1^2+2bx_1+2c~~~(3),~~~f(x_2)=ax_2^2+2bx_2+2c~~~(4)$$ Use (1) in (4) to get $f(x_1)=-ax_1^2$ and (2) in (4) to get $f(x_2)=3ax_2^2$ So $$f(x_1)f(x_2)=-3a^2x_1^2 x_2^2<0$$ Therefore by IVT there exists $x_3 \in (x_1,x_2)$ such that $f(x_3)=0.$