Quadratic equation and Proof

Question

For rational numbers $a$ and $b$, the quadratic equation $x^2 - ax - b = 0$ has two solutions according to my professor.

How can I Prove that if one of these is solutions is rational, the other must be as well. We have been given a small hint: if we like to use the linear factorization theorem in the proof.

Are you allowed to use the quadratic formula? If so, then one solution being rational should allow you to show the other is rational. What have you tried? — Kris Williams, May 06 '13 at 02:22

score 6 · Answer 1 · answered May 06 '13 at 02:24

6

If the roots are $r_1$ and $r_2$, by Vieta's formulas $a = r_1 + r_2$ and $b = - r_1 r_2$. Or just multiply $(x - r_1) (x - r_2)$.

answered May 06 '13 at 02:24

vonbrand

28,394

score 4 · Answer 2 · answered May 06 '13 at 02:25

We can do it the ugly way. The roots are $\frac{a\pm\sqrt{a^2+4b}}{2}$. Thus the sum of the roots is $a$. If one of the roots $r$ is rational, so is the other, which is $a-r$, since the difference of rationals is rational.

Remark: There are nicer ways to show that the sum of the roots is $a$. But I wanted to show that calculation will do the job.

Quadratic equation and Proof

2 Answers2