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Math Help - need help ASAP. my exams 2morow

  1. #1
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    Unhappy need help ASAP. my exams 2morow

    the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?
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  2. #2
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    Quote Originally Posted by outlaw View Post
    the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?
    Do you mean
    px^2 + 6x + q = 0
    has roots x = -3?

    Look at the discriminant of the quadratic:
    D = 6^2 - 4pq = 36 - 4pq

    If the two roots of a quadratic equation are equal, then we know the discriminant is 0:
    36 - 4pq = 0 =\implies q = \frac{9}{p}

    So by the quadratic formula, the solution to px^2 + 6x + q = 0 will be:
    x = \frac{-6 \pm D^2}{2p} = -3

    But D = 0, so
    \frac{-6}{2p} = -3

    -6 = -6p

    p = 1

    Thus q = \frac{9}{1} = 9.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by outlaw View Post
    the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?
    us ethe quadratic formula.

    the roots of the equation y = ax^2 + bx + c are given by:

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

    now for equal roots, we require that b^2 - 4ac = 0

    now continue
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  4. #4
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    I was editing my post in the other thread to note that b^2-4ac=0, and then you have two equations of two unknowns p and q to solve for. Follow this thread.
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    If px^2+6x+q has one repeated root; x=-3

    Then \implies px^2+6x+q=(x+3)(x+3)
    \implies px^2+6x+q=x^2+6x+9

    Compare co-effs.
    Last edited by Sean12345; January 7th 2008 at 01:07 PM. Reason: No MATH Tags
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sean12345 View Post
    If px^2+6x+q has one repeated root; x=-3

    Then \implies px^2+6x+q=(x+3)(x+3)
    \implies px^2+6x+q=x^2+6x+9

    Compare co-effs.
    yup, that's a lot easier. classic, thanks for pointing it out
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  7. #7
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    Quote Originally Posted by Sean12345 View Post
    If px^2+6x+q has one repeated root; x=-3

    Then \implies px^2+6x+q=(x+3)(x+3)
    \implies px^2+6x+q=x^2+6x+9

    Compare co-effs.
    This is easier but unfortunately it's also not true in general.

    The correct implication is px^2+6x+q=a(x+3)(x+3).

    It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    This is easier but unfortunately it's also not true in general.

    The correct implication is px^2+6x+q=a(x+3)(x+3).

    It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.
    so true. this caught me in a thread a while back. i realized i needed a constant multiplier when back-substituting. don't remember what thread it was
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    This is easier but unfortunately it's also not true in general.

    The correct implication is px^2+6x+q=a(x+3)(x+3).

    It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.
    True, i used that method because the 6x shows that a is equal to 1 hence
    px^2+6x+q=1(x+3)(x+3)
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