some more Algebra

how is

y = sqrt(2x-x^2) solved for x in terms of y?

I'm getting lost on this one

here is what i come up with

x = {[sqrt(-1)(sqrt(y^2 +1))]/-1} + 1

heres the steps i took if anyone get lost in that mess

y^2 = -x^2 + 2x

y^2 = - ( x^2 -2x)

y^2 + 1 = -( x^2 - 2x + 1 )

y^2 + 1 = -(x-1)^2

(y^2 + 1)/-1 = (x-1)^2

sqrt[ (y^2 + 1)/-1] = x-1

{sqrt(-1)[sqrt(y^2 + 1)/-1] +1 = x

Y^2 = 2X - X^2

-Y^2 = x^2 - 2X

1 - Y^2 = X^2 - 2X + 1

1 - Y^2 = (X - 1)^2

Sqrt (1 - Y^2) = X - 1

1 + sqrt (1 - Y^2) = X

smomofo has it right - this is a method called "completing the square." essentially the goal is to add the same constant to both sides so as to turn the quadratic in x into a perfect square.

ex: [some function of y] = x^2 + 4x

add 4 to both sides: [some function of y] + 4 = x^2 +4x +4 = (x+2)^2

now you can sqrt both sides: sqrt( [function of y] +4) = x+2

and subtract 2, and rearrange: x = sqrt([function of y]+4) - 2

in general, if the square you are completing is x^2 + cx then you will want to add (c/2)^2 to both sides, and the quadratic can be rewritten (x+c/2)^2.

note: i just looked back at your algebra and noticed that your completing of the square looked ok - you seem to understand the idea.

however, when you had y^2 = -(x^2-2x) you added 1 to the left, and added 1 to the right INSIDE the parentheses, which has a -1 coefficient. essentially, this is like adding one to the left side, and subtracting one from the right side. be careful!

first divide both sides by -1 so the x^2 term has +1 as its coefficient, THEN add 1 to both sides. this will give you the proper 1-y^2 on the left, or possibly written, -y^2 + 1 (which is equally correct)

good luck!