Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.

Question

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}

Answer 1

If $a=b=0$ and the root is $0$.

However, if $(a,b) \neq 0$, then

the discriminant is equal to $-4(a^4+4b^4)<0$, hence the roots are imaginary.

Answer 2

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!