Question

Let P(x) = x² + a + b be a quadratic polynomial with real coefficient. Suppose there are real numbers s ≠ t such that P(s) = t and P(t) = s. Prove that b-st is a root of the equation x² + a + b -st = 0.

Answer 1

Simplification gives st = 1 + a + b = P(I). This shows that  x = 1 is a root of x² + ax + b - st = 0. Since the product of root is b - st, the other root  is b - st.