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March 6, 2015

March 6, 2015

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

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 8:10pmyou 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 8:16pmim confused

- re:math -
**bobpursley**, Wednesday, February 24, 2010 at 9:09pmCheck with your teacher, then.

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