2018
Problem - 4763
Let $a$, $b$, $c$ and $d$ be real numbers. Find the relation of these four numbers such that the two curves $y=ax^2+c$ and $y=bx^2 + d$ have exactly two points of intersections.
This is equivalent to the function $(a-b)x^2 + (c-d)=0$ has two distinct real solutions. This means that $a-b\ne 0$. Hence we have $x^2=-\frac{c-d}{a-b}$.This equation has two distinct real root if and only if $$-\frac{c-d}{a-b} > 0\Leftrightarrow (a-b)(c-d) < 0$$