hi this is my first thread, im doing corrections for a math test i bombed i have the equation x^2+y^2+2x-6y-6=0 can someone help me find the center and the radius
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Originally Posted by osoares10 hi this is my first thread, im doing corrections for a math test i bombed i have the equation x^2+y^2+2x-6y-6=0 can someone help me find the center and the radius Complete the square in x-terms and y-terms and re-arrange to get the form $\displaystyle (x - h)^2 + (y-k)^2 = r^2$.
can u solve it too for me
Originally Posted by osoares10 can u solve it too for me No. I have told you what to do. Make an attempt. Show all your working. Say where you get stuck.
ok i get down to x^2+2x___+y^2-6y___=6 i cant find the numbers that go in the blank
Originally Posted by osoares10 ok i get down to x^2+2x___+y^2-6y___=6 i cant find the numbers that go in the blank The x-terms can be written (x + 1)^2 - 1. You do the y-terms. It looks like you need to thoroughly review how to complete the square.
Originally Posted by mr fantastic The x-terms can be written (x + 1)^2 - 1. You do the y-terms. It looks like you need to thoroughly review how to complete the square. isnt 1^2 = 1? please i really need to know this if i see it i will, but i can only find examples with a perfect square, i just cant figure it out
Last edited by mr fantastic; Oct 12th 2009 at 04:04 PM. Reason: Merged posts
Originally Posted by osoares10 isnt 1^2 = 1? please i really need to know this if i see it i will, but i can only find examples with a perfect square, i just cant figure it out Read this: Completing the Square: Solving Quadratic Equations
can some one please just do this for me, i just cant get it, i hate my teacher, please its very important
i think i got it, is the center -1,3 and the radius 4?
Originally Posted by osoares10 i think i got it, is the center -1,3 and the radius 4? Yes. Well done.
Originally Posted by mr fantastic Yes. Well done. thanks alot for the help
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