Number of values of x satisfying the pair of quadratic equations $x^2-px+20$ and $x^2-20x+p$ for $p\in R$

Question

Subtracting both equations $$x(20-p)+20-p=0$$ $$x=-1$$ That’s one root, but the answer is three. Where do the the other two roots come from?

Edit- I tried using the common roots equation, and I got the values of p as 20 and -21. But only 2 works while -21 gives two different roots for both the equations. Why is this happening?

Answer 1

If $x$ satisfies both equations, it is easy to see that $$ (p - 20) x - (20-p) = 0$$ then $$ (p - 20) (x + 1) =0 $$. If $p = 20$ then all values of $x$ solve the problem. If $p \neq 20$ then only possible value is $x = -1$.