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Math Help - another question about polynomials

  1. #1
    Junior Member
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    this is another question.

    what i tried to do is :

    x=b,

    (b^2) + pb +q = 3(b^2) +q
    -2(b^2) + pb = 0

    and then.....i'm stuck. ="=
    i think my beginning is already incorrect. =_=
    Last edited by mr fantastic; December 29th 2008 at 11:50 AM. Reason: Created a new thread for a new question added as an edit to an old post and tidied things up.
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  2. #2
    Lord of certain Rings
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    A tip: Always use new threads for new questions. Your problem will get attention faster. People always tend to answer new posts than old ones.

    Quote Originally Posted by wintersoltice View Post

    (b^2) + pb +q = 3(b^2) +q
    -2(b^2) + pb = 0
    Instead of the above equation, write it as:

    (b^2) + pb +q = 0; --------------------------(1)
    3(b^2) +q = 0;--------------------------(2)

    Multiply the first equation by 3 and subtract the second from it.

    (3b^2 + 3pb + 3q) - (3b^2 +q ) = 0  \Rightarrow 3pb + 2q = 0 \Rightarrow   9p^2b^2 = 4q^2

    Now eliminate b^2 from the above equation and equation (2).
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