prove for p(x) which is a quadratic polynomial

Question

$p(x)$ is a quadratic polynomial . Prove that any given number for $a$ with one exception , we can find a number $b$ such that $p(a)=p(b)$ and $a$ is not equal to $b$.

Answer 1

p(x) = mx^2 + nx + p, with m not equal to 0. Take p(b) - p(a) = m(b^2 - a^2) + n(b - a) = (b - a)(mb + ma + n). So given a, to get p(b) = p(a) choose b = -(ma +n)/m, then b is not equal a with the exception that a is not equal -n/2m which is the vertex of the parabola defined by p.

Answer 2

Hint: By completing the square, we can express any quadratic polynomial $F(x)$ in the form $$F(x)=p\left((x-q)^2+r\right)$$ where $p\ne 0$.