My Math Forum Hi guys, Solving for the inverse, How do you approach these?

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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 $y^2 - y = y^2 - y + 1/4 - 1/4 = (y-1/2)^2 - 1/4$ 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^2-y)=x completing the square => ((1/2)(-1))^2 = (1/4) (y^2 - y + 1/4) = x + 1/4 (y-1/2)^2 = x + 1/4 y = +- sqrt(x + 1/4) + 1/2 Thank!! This forum is the best kept secret

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