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October 4th, 2013, 04:02 PM  #1 
Member Joined: Aug 2013 Posts: 69 Thanks: 0  Hi guys, Solving for the inverse, How do you approach these?
y = x^2  x ....you have to complete the square? for x = y^2  y... how do you think these through? Zen. 
October 4th, 2013, 07:08 PM  #2 
Newbie Joined: Oct 2013 Posts: 10 Thanks: 0  Re: Hi guys, Solving for the inverse, How do you approach th
You want to isolate for y. But how do you take those two y terms and merge them into one nice y = .... term? The approach here would be to turn your y^2  y expression into something that looks like (y + a)^2. (y+a)^2 would produce a y^2 term. What a would produce your y term? After some thinking I think a = 1/2 would work But, (y  1/2)^2 = y^2  y + 1/4. That's not quite what we want, so let's remove that 1/4 term So From here you can isolate y. 
October 4th, 2013, 07:09 PM  #3 
Newbie Joined: Oct 2013 Posts: 10 Thanks: 0  Re: Hi guys, Solving for the inverse, How do you approach th
^ should be 1/2, not 1/2

October 5th, 2013, 12:45 AM  #4 
Member Joined: Aug 2013 Posts: 69 Thanks: 0  Re: Hi guys, Solving for the inverse, How do you approach th
Ah, thanks, yah, reminded me...completing the square. (y^2y)=x completing the square => ((1/2)(1))^2 = (1/4) (y^2  y + 1/4) = x + 1/4 (y1/2)^2 = x + 1/4 y = + sqrt(x + 1/4) + 1/2 Thank!! This forum is the best kept secret 

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