solve for x please

**ahmedb** · November 15th 2011, 06:36 PM

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

**pickslides** · November 15th 2011, 06:38 PM

Take 10-x from both sides, then square both sides. Does that make sense?

**Amer** · November 15th 2011, 06:39 PM

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

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

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

**ahmedb** · November 15th 2011, 06:45 PM

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

**pickslides** · November 15th 2011, 06:49 PM

Yes, expand the LHS then group like terms.

**ahmedb** · November 15th 2011, 06:54 PM

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)

**pickslides** · November 15th 2011, 07:28 PM

Try $(x+5.196)(x+5.196)=2(18+2x^2)$

Thread: solve for x please