How to prove a solution of equation is rational if another one is rational number?

Question

The question is : $r$ is the solution of equation $x^2+bx+c=0$ and $r$ is a rational number, so there is another solution $s$, how to prove s is a rational number as well? I have no idea about it and I think that I shall prove it through $(x-r)(x-s)$ equals to the equation, but still no idea about it. Please help me a little bit. Thanks.

Answer 1

Since $r$ is rational, say $r=p/q$, where $p,q\in\mathbb{N}$, and $p,q$ are relatively prime.

Then if $s$ is also a root, we have $x^2+bx+c=k(x-r)(x-s)\implies ax^2+bx+c=x^2-(r+s)x+rs$.

Thus, $$-(r+s)=b$$ $$rs=c$$

Using the last equation, we see that $s=c/r=\frac{cq}{p}$. So $s$ is only rational if $c$ is rational.

Answer 2

Well, there is a very easy way of going about this.

$r$ must be in the form $\frac uv$.

If $x=r$ is the solution to $x^2+bx+c$, then we have:

$$x=\frac{-b\pm\sqrt{b^2-4c}}2$$

Which we know one of such must be rational.

Lets go along with the following:

$$r=\frac{-b+\sqrt{b^2-4c}}2$$

We know that since $r$ is rational, then it is in the form $\frac uv$:

$$\frac uv=\frac{-b+\sqrt{b^2-4c}}2$$ We can easily see that $\sqrt{b^2-4c}$ must be rational for the numerator to be rational, which must be true because the denominator is rational and the whole fraction must be rational.

If $\sqrt{b^2-4c}$ is rational, then:

$$\frac{-b-\sqrt{b^2-4c}}2=\frac{-b-R_0}2=\frac{R_1}2=R_2$$

Where $R$ represents a rational number.

So both roots must be rational.