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

• Jan 20th 2009, 07:13 PM
wil
proof of that one rational solution to a quadratic implies another
Quote:

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! :)
• Jan 20th 2009, 07:16 PM
Mathstud28
Quote:

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.
• Jan 20th 2009, 10:41 PM
Constatine11
Quote:

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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• Jan 20th 2009, 10:45 PM
Constatine11
Quote:

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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• Jan 20th 2009, 10:48 PM
Constatine11
Quote:

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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• Jan 21st 2009, 12:34 PM
Mathstud28
Quote:

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? (Worried)
• Jan 21st 2009, 02:13 PM
Constatine11
Quote:

Originally Posted by Mathstud28
I'm sorry, was there a problem with my solution? (Worried)

No, just with my reading of it.

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