find x: 15.196=2*sqrt(3^2+x^2) + 10-x
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Take 10-x from both sides, then square both sides. Does that make sense?
$\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 $
Originally Posted by Amer $\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 $ so do i do this? (x+5.196)(x+5.196)=18+2x^2
Yes, expand the LHS then group like terms.
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)
Try $\displaystyle (x+5.196)(x+5.196)=2(18+2x^2)$
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