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Mathematics 23 Online
OpenStudy (anonymous):

solve for x. y=x^2+2x-1

OpenStudy (anonymous):

Gonna have to complete the square.

OpenStudy (anonymous):

good idea

OpenStudy (anonymous):

Or plug y into the quadratic formula.. But I prefer the former.

OpenStudy (anonymous):

me too. especially since you have 2x

OpenStudy (anonymous):

yep

OpenStudy (vijay):

x^2+2x-(1 + y) = 0

OpenStudy (anonymous):

ew.. vijay's doin in the other way.

OpenStudy (anonymous):

\(y = x^2 + 2x + 0 - 1\) \[\implies y = x^2 + 2x + (1 - 1) - 1\]\[\implies y = x^2 + 2x + 1 - 2\]\[\implies y = (x+1)^2 - 2\]\[\implies y + 2 = (x+1)^2\]\[\implies x + 1 = \sqrt{y+2}\]\[\implies x = \sqrt{y+2} - 1\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Oh, forgot the +-

OpenStudy (anonymous):

that's ok i put it in

OpenStudy (anonymous):

Yep

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