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Math Help - Quadratic equation

  1. #1
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    Quadratic equation

    a problem im having problems with...
    consider the quadratic equation (-2p+1)x^2 + (p-2)x + 6p = 0
    A) find the discriminant (7p-2)^2?
    B) for p not equal to 1/2 show that there are always two rational solutions and find these solutions.
    Could someone check if A is right and I dont know how to work out B?
    thanks *
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by needhelpplease View Post
    a problem im having problems with...
    consider the quadratic equation (-2p+1)x^2 + (p-2)x + 6p = 0
    A) find the discriminant (7p-2)^2?
    B) for p not equal to 1/2 show that there are always two rational solutions and find these solutions.
    Could someone check if A is right and I dont know how to work out B?
    thanks *
    The discriminat is

    b^2-4ac=(p-2)^2-4(-2p+1)(6p)=p^2-4p+4+48p^2-24p=49p^2-28p+4=(7p-2)^2

    correct

    so by the quadratic formula the solutions are

    x=\frac{(2-p) \pm (7p-2)}{2-4p} this will have two distinct rational solutions unless

    7p-2=0 if that is the case you will only have one reapeated rational solution.

    7p-2=0 \iff p=\frac{2}{7}

    I don't know where they got the one half from....
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  3. #3
    Junior Member mathhomework's Avatar
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    Please check my answer!

    a. your answer is correct, because the discriminant=b^2-4ac, which is
    (p-2)^2-4(-2p+1)(6p)=49p^2-28p+4=(7p-2)^2

    b. use quadratic forumular to solve.

    -b+/-sqrt(7p)^2)/2a

    =-(p-2)+/-sqrt(7p-2)^2/2(-2p+1)

    =2, 3p/1-2p
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