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Math Help - Discrete Proof! HELP!

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    Exclamation Discrete Proof! HELP!

    Prove that if one root of ax^2+bx+c=0 is rational, then both are rational.
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    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.
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    <br />
x = \frac {-b \pm \sqrt{b^2-4 \cdot a \cdot c }}{2 \cdot a}<br />
    if  \sqrt{b^2-4 \cdot a \cdot c } is rational then both root are rational.
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    How do I show that the other root is rational?
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    Quote Originally Posted by megster_567 View Post
    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,
    <br />
\frac{{ - b - \sqrt {b^2  - 4ac} }}{{2a}} mean?
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  6. #6
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    Quote Originally Posted by Plato View Post
    One root of this equation is \frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}} now if that is rational then what does the other root,
    <br />
\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...
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