SOLUTION: Hi, I solved this quadratic equation but was not sure how to finalize my answer. I thought it came out to be a "not a real solution" but then I pulled out the -6 and continued. W

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Question 409987: Hi, I solved this quadratic equation but was not sure how to finalize my answer. I thought it came out to be a "not a real solution" but then I pulled out the -6 and continued. We have not learned how to write it as a complex number yet, so we write it as "not a real solution". What do you think?
(x-3)^2=-9
(x-3)(x-3)+9=0
x^2-3x-3x+9+9=0
x^2-6+18=0
-6+-sqrt((-6)^2-4(1)(18))/2(1)
-6+-sqrt(36-72)/(2)
-6+-sqrt(-36)/(2)
(-6+-(-6))/(2)
x= -(6)/1 which is -6 and x = 0
or for the first x answer do I write it as:
6+-6sqrt-1

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
(x-3)^2=-9
(x-3)(x-3)+9=0
x^2-3x-3x+9+9=0
x^2-6x+18=0
x = [-6+-sqrt((-6)^2-4(1)(18))]/2(1)
x = [-6+-sqrt(36-72)]/(2)
x = [-6+-sqrt(-36)]/(2)
---
I changed your next step because sqrt(-36) is 6i
---
x = [(-6+-(6)i)/(2)
---
2 is a common denominator so you get:
--------------
x = -3+3i or x = -3-3i
============================
Cheers,
Stan H.

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