1. ## Discrete Proof! HELP!

Prove that if one root of $ax^2+bx+c=0$ is rational, then both are rational.

2. You know that solution of a quadratic equation follows the quadratic formula. Then assume that one of the roots is rational, and not complex, so sqrt(b^2-4ac) is a rational number. Then show that the other root must also be rational.

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

if $\sqrt{b^2-4 \cdot a \cdot c }$ is rational then both root are rational.

4. ## Thanks

How do I show that the other root is rational?

5. Originally Posted by megster_567
Prove that if one root of $ax^2+bx+c=0$ is rational, then both are rational.
One root of this equation is $\frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}}$ now if that is rational then what does the other root,
$
\frac{{ - b - \sqrt {b^2 - 4ac} }}{{2a}}$
mean?

6. Originally Posted by Plato
One root of this equation is $\frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}}$ now if that is rational then what does the other root,
$
\frac{{ - b - \sqrt {b^2 - 4ac} }}{{2a}}$
mean?

I think I got that, I'm just not sure how to incorporate it into a proof...