If the roots of $a{x}^{2}+bx+c$, where $a$ and $c$ are non-zero real numbers, be imaginary and the ratio of the roots be $r:1$, where $r>0$, then prove that $a$ and $c$ are of opposite signs.
My approach: For roots being imaginary, we get ${b}^{2} - 4{a}{c}< 0$ (First condition)
Let the roots be $\alpha$ and $\beta$ respectively. We know that $\alpha +\beta =\frac { -b }{ a } $ and $\alpha \beta =\frac { c }{ a }$.
Squaring the sum expression and dividing with the product's , we get:
$\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +2=\frac { { b }^{ 2 } }{ ac } <4$ or $r+\frac { 1 }{ r } <2$.
As AM-GM is applicable for non-negative real numbers, we get $r+\frac { 1 }{ r } \ge 2$.
As observable, the intersection of both the conditions is $\phi $.
Kindly let me know where I am going wrong. Any hint or solution could be appreciated.
$$ax^2+bx+c=a(x-(p+qi))(x-r(p+qi)),$$ where $\{p,q,\}\subset\mathbb R$.
Thus, we need to prove that $$(p+qi)^2<0,$$ which is $pq=0$ and $p^2-q^2<0$, which is obvious because roots are imaginary.