fastlikearabbit wrote: Let $a$ and $b$ be two real numbers. Find these numbers given that the graphs of $f:\mathbb{R} \to \mathbb{R} , f(x)=2x^4-a^2x^2+b-1$ and $g:\mathbb{R} \to \mathbb{R} ,g(x)=2ax^3-1$ have exactly two points of intersection. Let $h(x)=f(x)-g(x)=2x^4-2ax^3-a^2x^2+b$ $h'(x)=8x^3-6ax^2-2a^2x$ $=2x(x-a)(4x+a)$ $h(0)=b$ $h(a)=b-a^4$ $h(-\frac a4)=b-\frac{3a^4}{128}$ And so we need : Either $b<0$ Either $b>0$ and $b-a^4<0<b-\frac{3a^4}{128}$ Hence the answer : $\boxed{b\in(-\infty,0)\cup\left(\frac{3a^4}{128},a^4\right)}$

Compute the greatest real number $K$ for which the graphs of $$(|x| - 5)^2 + (|y| - 5)^2 = K \hspace{3mm} \text{and} \hspace{3mm} (x - 1)^2 + (y + 1)^2 = 37$$have exactly two intersection points. (ARML 2019)