proof of that one rational solution to a quadratic implies another

**wil** · January 20th 2009, 07:13 PM

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 $x^{2}+bx+c=(x-r)(x-s)$ .

For my proof, I'm hypothesizing that $s$ 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 $x$ is rational, then that closure ensures a rational $r$ , but $x$ 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 $x$ hurdle?

Thanks!

**Mathstud28** · January 20th 2009, 07:16 PM

Originally Posted by wil

For my proof, I'm hypothesizing that $s$ 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 $x$ is rational, then that closure ensures a rational $r$ , but $x$ 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 $x$ hurdle?

Thanks!

Try comparing coefficients

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

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

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

**Constatine11** · January 20th 2009, 10:41 PM

Originally Posted by Mathstud28

Try comparing coefficients

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

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

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

.

**Constatine11** · January 20th 2009, 10:45 PM

Originally Posted by Mathstud28

Try comparing coefficients

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

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

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
.

**Constatine11** · January 20th 2009, 10:48 PM

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

.

**Mathstud28** · January 21st 2009, 12:34 PM

Originally Posted by Constatine11

Too quick, post edited, and for some reason duplicated

.

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

**Constatine11** · January 21st 2009, 02:13 PM

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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