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October 4th, 2013, 04:02 PM   #1
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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.
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October 4th, 2013, 07:08 PM   #2
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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.
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October 4th, 2013, 07:09 PM   #3
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Re: Hi guys, Solving for the inverse, How do you approach th

^ should be -1/2, not 1/2
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October 5th, 2013, 12:45 AM   #4
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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!!

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