Solve \ I tried most things but get stuck at pretty much every time. Any help?
Last edited by Mukilab; June 5th 2010 at 12:53 PM.
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Originally Posted by Mukilab Solve I tried most things but get stuck at pretty much every time. Any help? You're fine up to there, you can complete the square to find y. Remember that
How can I complete the square with that!? I thought completing the square is along the lines of 2x^2+4x+13=(2x^2+4x+4)-11 o_o What can I do with {y-1}^2? y^2-2y+1 completing the square (y^2-2y+1)+1-1=9/4
Originally Posted by Mukilab How can I complete the square with that!? I thought completing the square is along the lines of 2x^2+4x+13=(2x^2+4x+4)-11 o_o What can I do with {y-1}^2? y^2-2y+1 completing the square (y^2-2y+1)+1-1=9/4 is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1.
Originally Posted by e^(i*pi) is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1. I've done that already but I got 2 and 4 which don't work if you put them into the equation
OOhhhhh Thanks, I forgot when I did it you have to take the square root on both sides! heh. Don't I need an exact answer though? Not allowed a calculate final:
Originally Posted by e^(i*pi) Remember that Originally Posted by Mukilab OOhhhhh Thanks, I forgot when I did it you have to take the square root on both sides! heh. Don't I need an exact answer though? Not allowed a calculate final: See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the sign. edit: that answer is correct, it just needs simplifying and also the negative solution
Originally Posted by e^(i*pi) See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the sign. edit: that answer is correct, it just needs simplifying and also the negative solution ???? sorry
Originally Posted by Mukilab That's 3/2 + 1 = 5/2
Originally Posted by Mukilab ???? sorry Don't ever be sorry for asking for help As that square root can be reduced You should get hence We can't add the 1 under the square root since
Originally Posted by e^(i*pi) Don't ever be sorry for asking for help As that square root can be reduced You should get hence We can't add the 1 under the square root since Thanks for your perserveerance
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