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solve for x. y=x^2+2x-1
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Gonna have to complete the square.
good idea
Or plug y into the quadratic formula.. But I prefer the former.
me too. especially since you have 2x
yep
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x^2+2x-(1 + y) = 0
ew.. vijay's doin in the other way.
\(y = x^2 + 2x + 0 - 1\) \[\implies y = x^2 + 2x + (1 - 1) - 1\]\[\implies y = x^2 + 2x + 1 - 2\]\[\implies y = (x+1)^2 - 2\]\[\implies y + 2 = (x+1)^2\]\[\implies x + 1 = \sqrt{y+2}\]\[\implies x = \sqrt{y+2} - 1\]
\[x^2+2x-1=y\] \[x^2+2x=y+1\] \[(x+1)^2=y+1+1\] \[(x+1)^2=y+2\] \[x+1=\pm\sqrt{y+2}\] \[x=-1\pm\sqrt{y+2}\]
Oh, forgot the +-
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that's ok i put it in
Yep
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