Solving for x. How?

• May 17th 2009, 05:22 PM
dwat
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

0 = 80x(500^2+x^2)^(-1/2) - 50
• May 17th 2009, 05:36 PM
skeeter
Quote:

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

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
• May 18th 2009, 01:51 AM
dwat
perfect, thanks. It is correct, I just wanted to clarify how the textbook got the answer.