<s>[INST] Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary. Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.
My Attempt:
$$x^2-4abx+(a^2+2b^2)^2=0$$
Comparing above equation with $Ax^2+Bx+C=0$, we get
\begin{align}
A&=1 \\
B&=-4ab \\
C&=(a^2+2b^2)^2
\end{align}
Now,
\begin{align}
B^2-4AC
&=(-4ab)^2 - 4\cdot 1\cdot (a^2+2b^2)^2 \\
&=16a^2b^2 - 4(a^4+4a^2b^2+4b^4) \\
&=-4a^4-16b^4. \end{align}
 [/INST] Suppose $ab\ne 0$.
Let $f (x) $ be your quadric.
$$\lim_{\infty}f (x)=+\infty $$
thus the roots will not be real if the minimum is $$f (c)>0$$ where $c $ is such that $$f'(c)=0.$$
$$f'(x)=2x-4ab \implies c=2ab $$
$$\implies f (c)=$$
$$4a^2b^2-8a^2b^2+(a^2+2b^2)^2=$$
$$(a^2+2b^2+2ab)(a^2+2b^2-2ab) =$$
$$((a+b)^2+b^2)((a-b)^2+b^2)>0$$
Done!
</s>