# Thread: proof of that one rational solution to a quadratic implies another

1. ## proof of that one rational solution to a quadratic implies another

Prove that if one solution for a quadratic equation of the form $\displaystyle x^{2}+bx+c=0$ is rational (where $\displaystyle b$ and $\displaystyle c$ are rational), then the other solution is also rational. (Use the fact that if the solutions of the equation are $\displaystyle r$ and $\displaystyle s$, then $\displaystyle x^{2}+bx+c=(x-r)(x-s)$.
For my proof, I'm hypothesizing that $\displaystyle s$ is the solution that is known to be rational.

So the rationals are closed under multiplication, addition and subtraction (and division, with nonzero divisors.) If $\displaystyle x$ is rational, then that closure ensures a rational $\displaystyle r$, but $\displaystyle x$ isn't necessarily rational. I know I'm missing something, but I've been doing math all day and I'm a little brain-burned. Can anyone nudge me past the whole non-rational $\displaystyle x$ hurdle?

Thanks!

2. Originally Posted by wil
For my proof, I'm hypothesizing that $\displaystyle s$ is the solution that is known to be rational.

So the rationals are closed under multiplication, addition and subtraction (and division, with nonzero divisors.) If $\displaystyle x$ is rational, then that closure ensures a rational $\displaystyle r$, but $\displaystyle x$ isn't necessarily rational. I know I'm missing something, but I've been doing math all day and I'm a little brain-burned. Can anyone nudge me past the whole non-rational $\displaystyle x$ hurdle?

Thanks!
Try comparing coefficients

$\displaystyle (x-s)(x-r)=x^2-rx-sx+rs=x^2-(r+s)x+rs=x^2+bx+c$

So $\displaystyle r+s=-b$ and $\displaystyle rs=c$

And since as you stated the rationals are closed under addition/multiplication the conclusion follows.

3. Originally Posted by Mathstud28
Try comparing coefficients

$\displaystyle (x-s)(x-r)=x^2-rx-sx+rs=x^2-(r+s)x+rs=x^2+bx+c$

So $\displaystyle r+s=-b$ and $\displaystyle rs=c$

And since as you stated the rationals are closed under addition/multiplication the conclusion follows.
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4. Originally Posted by Mathstud28
Try comparing coefficients

$\displaystyle (x-s)(x-r)=x^2-rx-sx+rs=x^2-(r+s)x+rs=x^2+bx+c$

So $\displaystyle r+s=-b$ and $\displaystyle rs=c$

And since as you stated the rationals are closed under addition/multiplication the conclusion follows.
You only need that the sum of the rational root and the other is a rational and closure under addition and subtraction
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5. Originally Posted by Jhevon
i think it is fine. since we are assuming either r or s is rational to begin with, if their sum is rational, then they both have to be rational, since the sum of a rational number and an irrational number would be irrational. so a proof by contradiction can work here.
Too quick, post edited, and for some reason duplicated

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6. Originally Posted by Constatine11
Too quick, post edited, and for some reason duplicated

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I'm sorry, was there a problem with my solution?

7. Originally Posted by Mathstud28
I'm sorry, was there a problem with my solution?
No, just with my reading of it.

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