find x:

15.196=2*sqrt(3^2+x^2) + 10-x

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- Nov 15th 2011, 06:36 PMahmedbsolve for x please
find x:

15.196=2*sqrt(3^2+x^2) + 10-x - Nov 15th 2011, 06:38 PMpickslidesRe: solve for x please
Take 10-x from both sides, then square both sides. Does that make sense?

- Nov 15th 2011, 06:39 PMAmerRe: solve for x please
$\displaystyle 15.196 = 2\sqrt{3^2+x^2} + 10 -x $

$\displaystyle x -10 + 15.196 = 2 \sqrt{3^2+x^2} $ square both sides

$\displaystyle (x + 5.196)^2 = \left( 2\sqrt{3^2+x^2}\right)^2 $ - Nov 15th 2011, 06:45 PMahmedbRe: solve for x please
- Nov 15th 2011, 06:49 PMpickslidesRe: solve for x please
Yes, expand the LHS then group like terms.

- Nov 15th 2011, 06:54 PMahmedbRe: solve for x please
is the right hand side proper?

cas i tried (2*sqrt(3^2+3^2))^2 and got 72

and 18+2(3)^2 is 36.

so is this the correct RHS: 2(18+2x^2) - Nov 15th 2011, 07:28 PMpickslidesRe: solve for x please
Try $\displaystyle (x+5.196)(x+5.196)=2(18+2x^2)$