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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- May 17th 2009, 05:22 PMdwatSolving 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 - May 17th 2009, 05:36 PMskeeter
$\displaystyle \frac{80x}{\sqrt{500^2+x^2}} = 50$

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

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

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

$\displaystyle 39x^2 = 25 \cdot 500^2$

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

check both solutions in the original equation - May 18th 2009, 01:51 AMdwat
perfect, thanks. It is correct, I just wanted to clarify how the textbook got the answer.