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Math Help - Polya Problems

  1. #1
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    Polya Problems

    Question ONE:

    Let 'a' and 'b' be integers neither wof which is equal to zero. Prove that the equation...
    a^2x^3 + b^2x^2 – 1 = 0

    ...has no integer roots.



    Question TWO:

    For which real values of 'a' do the equations...
    x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0
    ...have a common root?
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  2. #2
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    Quote Originally Posted by edwardteo.368 View Post
    Question ONE:

    Let 'a' and 'b' be integers neither wof which is equal to zero. Prove that the equation...
    a^2x^3 + b^2x^2 – 1 = 0

    ...has no integer roots.
    The rational root theorem says that the possible rational roots of this polynomial are of the form \pm \frac{\text{factors of 1}}{\text{factors of }a^2}.

    For integer roots, then, we only have the possibility of -1 and 1.

    If it is 1 then
    a^2 + b^2  - 1 = 0
    To which we cannot have a and b both be integers.

    If it is -1 then
    -a^2 + b^2 - 1 = 0
    To which we cannot have a and b both be integers.

    So there are no integer solutions to the equation.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by edwardteo.368 View Post
    Question TWO:

    For which real values of 'a' do the equations...
    x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0
    ...have a common root?
    Take a look at the second equation. This is a biquadratic, so we may solve it in terms of a:
    x = \pm \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}}
    where the two \pm operations do not depend on each other.

    If the two equations have a common root then one version of this (some combination of the \pm's) must be a solution to the first equation:

    (Hang on to yer hats kiddies! This's gonna be one wild ro-dee-oh! Yee-haaaa! )

    \left ( \pm \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} \right ) ^3 + a \left ( \pm \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} \right ) + 1 = 0

    \pm \left ( \frac{-a \pm \sqrt{a^2 - 4}}{2} \right ) \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} \pm a \left ( \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} \right ) + 1 = 0

    \left ( \pm \frac{-a \pm \sqrt{a^2 - 4}}{2} \pm a \right ) \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} + 1 = 0

    \pm \left ( \frac{a \pm \sqrt{a^2 - 4}}{2} \right ) \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} + 1 = 0

    \pm \sqrt{\frac{-a \pm \sqrt{a^2 - 4}}{2}} = -\frac{2}{a \pm \sqrt{a^2 - 4}}

    \frac{-a \pm \sqrt{a^2 - 4}}{2} = \frac{2}{a^2 \pm a \sqrt{a^2 - 4} - 2}

    (-a \pm \sqrt{a^2 - 4})(a^2 \pm a \sqrt{a^2 - 4} - 2) = 4

    -a^3 + 2a + a(a^2 - 4) \mp 2\sqrt{a^2 - 4} = 4

    \mp \sqrt{a^2 - 4} = a + 2

    a^2 - 4 = a^2 + 4a + 4

    a = -2

    So we have the result that
    x^3 - 2x + 1 = 0 and x^4 - 2x^2 + 1 = 0
    have a common root. This root is x = 1.

    -Dan
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  4. #4
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    Thanks alot mate.

    Though I do have one question... what is that inverted plus/minus sign?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by edwardteo.368 View Post
    Thanks alot mate.

    Though I do have one question... what is that inverted plus/minus sign?
    It is used when there are signs that need to be ordered. For expample when \pm a - b = \pm c and the two \pm's carry the same sign (ie they are not independent) then if we divide both sides by -1:
    \frac{\pm a - b}{-1} = \frac{\pm c}{-1}

    The coefficients of a and c must keep the same relative sign, so
    \mp a + b = \mp c

    -Dan
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