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For a quadratic function, there is one minimum/maximum (the proof requires calculus) and also it is either always convex or concave (prove is also calculus) it is continuous every where, hence, it can have a maximum of 2 roots.

Graph it.

If there is more than 2 roots, by Intermediate Value Theorem, it cannot be convex/concave everywhere. It will HAVE to have two intervals of increasing or decreasing. It can be easily proven that given any quadratic function f(x), if x = x0 is a minimum/maximum, and x=a != x0 is a root, then 2x0-a is also a root. It is still true that a = x0 as 2x0-x0=x0 implying it is the only root.

But the concept of min/max requires Calculus to prove existence.

So, this is Calculus, not algebra.

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Q: How many x intercepts can a quadratic function have?
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