Homework Help: re:math

Posted by larry on Wednesday, February 24, 2010 at 7:52pm.

i am having a minor difficulty
i've been stuck for over an hour and don't know what to do
i dont know is this right

? = "+" or "-"
# = number

y=3
(x-3)^2 + (y-4)^2 = 9

i substitute y

(x-3)^2 + (3-4)^2 = 9
(x-3)^2 + 1 = 9
x^2 - 6x + 9 + 1 = 9
x^2 - 6x + 10 = 9
x^2 - 6x + 1 = 0
(x ? #)(x ? #) = 0

i get to here and now i am stuck
can u correct me if i did something wrong

please and thank you

re:math - bobpursley, Wednesday, February 24, 2010 at 8:10pm
you are correct.
Use the quadratic formula

x= (-b+-sqrt (b^2-4ac))/2a

or do it this way:

(x-3)^1+1=9
(x-3)^2=8= 2*2^2
take the sqrt of each side
x-3=+-2*sqrt2

x= 3+sqrt2 or x=3-sqrt2
re:math - larry, Wednesday, February 24, 2010 at 8:16pm
im confused
re:math - bobpursley, Wednesday, February 24, 2010 at 9:09pm
Check with your teacher, then.

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