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Math Help - Solving for x. How?

  1. #1
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    Solving for x. How?

    I have an equation that I need to solve for x in which i have no clue of how to solve it

    Please Help if you can

    0 = 80x(500^2+x^2)^(-1/2) - 50
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  2. #2
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    Quote Originally Posted by dwat View Post
    I have an equation that I need to solve for x in which i have no clue of how to solve it

    Please Help if you can

    0 = 80x(500^2+x^2)^(-1/2) - 50
    \frac{80x}{\sqrt{500^2+x^2}} = 50

    \frac{8x}{5} = \sqrt{500^2+x^2}

    \frac{64x^2}{25} = 500^2 + x^2

    \frac{39x^2}{25} = 500^2

    39x^2 = 25 \cdot 500^2

    x = \pm \sqrt{\frac{25 \cdot 500^2}{39}}

    check both solutions in the original equation
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  3. #3
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    perfect, thanks. It is correct, I just wanted to clarify how the textbook got the answer.
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