Question Statement:-
If the roots of the equation $ax^2+2bx+c=0$ are real and distinct, then show that the roots of the equation $$(a+c)\left(ax^2+2bx+c\right)=2(ac-b^2)(x^2+1)$$ are non-real complex numbers and vice versa.
The long standard solution:-
As the standard solution what I did was to find $D_2$ (subscript $2$ referring to terms related to the second equation and subscript $1$ referring to terms related to the first equation) and $D_1$, which comes out to be $$D_2=-4(b^2-ac)\left[4b^2+(a-c)^2\right]=-4D_1\left[4b^2+(a-c)^2\right]$$
Now, if $4b^2+(a-c)^2=0$. Then, $b=0$ and $a=c$.
This leads to the first equation(i.e. $ax^2+2bx+c=0$) becoming $x^2+1=0$, this results into the equation having complex roots which was not supposed initially, so we get $$4b^2+(a-c)^2\gt0$$
$$\therefore \qquad\qquad\dfrac{D_2}{D_1}\lt 0\qquad\qquad$$
Hence, if one of the given quadratic equations has complex roots then the other one is bound to have a real root.
My deal with the question:-
I was trying to come up with a more intuitive solution on looking at the way the second equation was arranged, i.e. $(a+c)\left(ax^2+2bx+c\right)=2(ac-b^2)(x^2+1)$
So what I thought was that the RHS has no real roots and is always negative (if $D_1\gt0$) and LHS has two distinct real roots, though it was not very useful but still thought of this so that I could come up with something.
After this what was necessary for both to be equal at some values was that when the graph of $ax^2+2bx+c$ was below $x-$axis (i.e. when $x$ is b/w the roots of the equation) it should be equal to RHS at some point. I know it sounds rather obvious that's why I need your help.