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Math Help - need help please :)

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    6

    need help please :)

    What are the zeros of y-7=4x^2

    and Solve square root of 2x^2 - sqaure root of 2 = 0
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  2. #2
    Member
    Joined
    Oct 2008
    Posts
    147
    y-7 = 4x^2
    -7 = 4x^2
    Never true in the reals.
    \sqrt(2x^2) - \sqrt(2) =0
    (\sqrt(2x^2)-\sqrt(2))(\sqrt(2x^2)+\sqrt(2)) =0
    2x^2 -2 =0
    x^2 -1 = 0
    x = -1 \, or\,  1
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