Question

Prove that if one solution for a quadratic equation of the form x 2 + bx + c = 0 is rational (where b and c are rational), then the other solution is also rational. (use the fact that if the solutions of the equation are r and s, then x 2 + bx + c = (x − r)(x − s).)

Answer 1

$x^2+bx+c = (x-r)(x-s)\\ x^2+bx+c = x^2-(r+s)x+rs\\\\ b = -(r+s)\\ c = rs$

Say r is rational. Suppose for a second, that s is not. Then, r+s is irrational. But this contradicts the fact that b is rational.

So, if one root is rational, then the other root is also rational