# solve for x please

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• Nov 15th 2011, 07:36 PM
ahmedb
solve for x please
find x:
15.196=2*sqrt(3^2+x^2) + 10-x
• Nov 15th 2011, 07:38 PM
pickslides
Re: solve for x please
Take 10-x from both sides, then square both sides. Does that make sense?
• Nov 15th 2011, 07:39 PM
Amer
Re: solve for x please
$15.196 = 2\sqrt{3^2+x^2} + 10 -x$

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

$(x + 5.196)^2 = \left( 2\sqrt{3^2+x^2}\right)^2$
• Nov 15th 2011, 07:45 PM
ahmedb
Re: solve for x please
Quote:

Originally Posted by Amer
$15.196 = 2\sqrt{3^2+x^2} + 10 -x$

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

$(x + 5.196)^2 = \left( 2\sqrt{3^2+x^2}\right)^2$

so do i do this?
(x+5.196)(x+5.196)=18+2x^2
• Nov 15th 2011, 07:49 PM
pickslides
Re: solve for x please
Yes, expand the LHS then group like terms.
• Nov 15th 2011, 07:54 PM
ahmedb
Re: 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, 08:28 PM
pickslides
Re: solve for x please
Try $(x+5.196)(x+5.196)=2(18+2x^2)$