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Math Help - Another quadratics!

  1. #1
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    Another quadratics!

    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
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    Quote Originally Posted by Gigabytes View Post
    [1] What are the zeros of y-7=4x^2

    and [2] Solve square root of 2x^2 - sqaure root of 2 = 0
    to [1].
    If I understand this question correctly then y = 0. Therefore solve for x:

    -7 = 4x^2~\implies~x^2=-\dfrac74~\implies~x=\pm i \cdot \dfrac12 \sqrt{7}

    to [2]:

    \sqrt{2x^2} - \sqrt{2} =0~\implies~\sqrt{2} \cdot \sqrt{x^2} = \sqrt{2} \implies~|x| = 1
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