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Thread: 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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    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
    $\displaystyle px^2 + 6x + q = 0$
    has roots x = -3?

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

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

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

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

    $\displaystyle -6 = -6p$

    $\displaystyle p = 1$

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

    -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 $\displaystyle y = ax^2 + bx + c$ are given by:

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

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

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

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

    Compare co-effs.
    Last edited by Sean12345; Jan 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 $\displaystyle px^2+6x+q$ has one repeated root; $\displaystyle x=-3$

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

    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 $\displaystyle px^2+6x+q$ has one repeated root; $\displaystyle x=-3$

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

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

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

    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 $\displaystyle px^2+6x+q=a(x+3)(x+3)$.

    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 $\displaystyle px^2+6x+q=a(x+3)(x+3)$.

    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
    $\displaystyle px^2+6x+q=1(x+3)(x+3)$
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