B behindDkeys May 2010 18 0 May 16, 2010 #1 as i return from vacation i forgot how to do this and i kinda lost my notes "Solve for the solution of:" \(\displaystyle y=x^2, y=8-x^2 \) please show me step by step

as i return from vacation i forgot how to do this and i kinda lost my notes "Solve for the solution of:" \(\displaystyle y=x^2, y=8-x^2 \) please show me step by step

e^(i*pi) MHF Hall of Honor Feb 2009 3,053 1,333 West Midlands, England May 16, 2010 #2 behindDkeys said: as i return from vacation i forgot how to do this and i kinda lost my notes "Solve for the solution of:" \(\displaystyle y=x^2, y=8-x^2 \) please show me step by step Click to expand... Set them equal and solve for x \(\displaystyle x^2 = 8-x^2\) which cancels down to \(\displaystyle x^2-4 = 0\) Solve using the difference of two squares Reactions: behindDkeys

behindDkeys said: as i return from vacation i forgot how to do this and i kinda lost my notes "Solve for the solution of:" \(\displaystyle y=x^2, y=8-x^2 \) please show me step by step Click to expand... Set them equal and solve for x \(\displaystyle x^2 = 8-x^2\) which cancels down to \(\displaystyle x^2-4 = 0\) Solve using the difference of two squares

B behindDkeys May 2010 18 0 May 16, 2010 #3 e^(i*pi) said: Set them equal and solve for x \(\displaystyle x^2 = 8-x^2\) which cancels down to \(\displaystyle x^2-4 = 0\) Solve using the difference of two squares Click to expand... oh yea and im gonna have answer of -2 and positive 2 now i remember(Rofl) (Clapping)thanks Last edited: May 16, 2010

e^(i*pi) said: Set them equal and solve for x \(\displaystyle x^2 = 8-x^2\) which cancels down to \(\displaystyle x^2-4 = 0\) Solve using the difference of two squares Click to expand... oh yea and im gonna have answer of -2 and positive 2 now i remember(Rofl) (Clapping)thanks