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Math Help - proof of that one rational solution to a quadratic implies another

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

    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!
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wil View Post
    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.
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    Quote Originally Posted by Mathstud28 View Post
    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.
    .
    Last edited by Constatine11; January 20th 2009 at 10:46 PM. Reason: double post for some reason
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    Quote Originally Posted by Mathstud28 View Post
    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
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    Quote Originally Posted by Jhevon View Post
    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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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Constatine11 View Post
    Too quick, post edited, and for some reason duplicated

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

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