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Math Help - Quadratic Identities - I Need Help Please.

  1. #1
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    Quadratic Identities - I Need Help Please.

    Find the Values of m, p and q for which 2x^2 - x -1 is equivilent to m(x+1)^2 + p(x+1) + q.

    I don't know how to show this. I have Tried to expand the Question but i could not come to anything rational.

    In this type of question do i use the quadratic formula?
    <br />
= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br />

    If so then.


    Line 1:
     \frac{-(-1) \pm \sqrt{-1^2 - 4*2*-1}}{2*-1} is equivilent to \frac{-(px+p) \pm \sqrt{(px-p)^2 - 4*m(x+1)*q}}{2*m(x+1)^2}

    Line 2:
     \frac{+1 \pm \sqrt{1 + 8}}{-2} is equivilent to \frac{-px-p \pm \sqrt{px^2 - 2p^2 x - 4mqx - 4mq}}{2mx^2 + 4mx + 2m}

    Line 3:
     \frac{+1 \pm 3}{-2} is equivilent to \frac{-px-p \pm \sqrt{px^2 - 2p^2 x - 4mqx - 4mq}}{2mx^2 + 4mx + 2m}

    This is where i am stuck.
    I know to find the values for m, p and q. I need to do simutanious equations. But i do not know where to apply it. I also think that using the Quadratic formula is also incorrect.

    Any help is appreciated. An explanation is even better.
    Thank you for any help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by name101 View Post
    Find the Values of m, p and q for which 2x^2 - x -1 is equivilent to m(x+1)^2 + p(x+1) + q.
    2x^2 - x - 1 = m(x + 1)^2 + p(x + 1) + q

    2x^2 - x - 1 = mx^2 + 2mx + m + px + p + q

    2x^2 - x - 1 = mx^2 + (2m + p)x + (m + p + q)

    Thus
    m = 2
    2m + p = -1
    m + p + q = -1

    You can solve for p and q.

    -Dan
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    m = 2
    2m + p = -1
    m + p + q = -1

    then
    2 * 2 + p = -1
    4 + p = -1
    p = -5

    m + p + q = -1
    2 + (-5) + q = -1
    -3 + q = -1
    q = 2

    that part was easy. for some reason that made a lot more sence.
    thanks

    so i take it the point of expanding out is to get
    mx^2 + b + c = 0
    even though it is expressed as
    mx^2 + (2m + p)x + (m + p + q) = 0
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by name101 View Post
    m = 2
    2m + p = -1
    m + p + q = -1

    then
    2 * 2 + p = -1
    4 + p = -1
    p = -5

    m + p + q = -1
    2 + (-5) + q = -1
    -3 + q = -1
    q = 2

    that part was easy. for some reason that made a lot more sence.
    thanks

    so i take it the point of expanding out is to get
    mx^2 + b + c = 0
    even though it is expressed as
    mx^2 + (2m + p)x + (m + p + q) = 0
    Yup. Whenever we have two polynomial expressions equal to each other the coefficients of like terms must be equal to each other.

    -Dan
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    Find the value m in x^2 + 2mx - 6 = 0 if one of the roots is 2.

    i know the answer is m = 0.5
    But i dont understand why.
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    Edit
    Last edited by name101; August 9th 2007 at 04:37 AM.
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  7. #7
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    Quote Originally Posted by name101 View Post
    Find the value m in x^2 + 2mx - 6 = 0 if one of the roots is 2.

    i know the answer is m = 0.5
    But i dont understand why.
    Let a = the other root, so,
    x^2 +2mx -6 = (x-2)(x-a)
    x^2 +2mx -6 = x^2 -ax -2x +2a
    x^2 +2mx -6 = x^2 -(a+2)x +2a

    So, 2a = -6
    a = -3

    And, -(a+2) = 2m
    So, -(-3 +2) = 2m
    1 = 2m
    m = 1/2 = 0.5 ----------------answer.
    Last edited by ticbol; August 9th 2007 at 12:32 PM.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by name101 View Post
    Find the value m in x^2 + 2mx - 6 = 0 if one of the roots is 2.

    i know the answer is m = 0.5
    But i dont understand why.
    Please post new questions in a new thread.

    -Dan
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  9. #9
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    sorry i thought since it is related. It can go in the same thread.
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