SOLUTION: Ok, so I am asked to write a quadratic equation based on the answer given. I have already done it, but someone is telling me I am wrong, and when I did it again, I came up with a

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Ok, so I am asked to write a quadratic equation based on the answer given. I have already done it, but someone is telling me I am wrong, and when I did it again, I came up with a       Log On


   

Question 223641: Ok, so I am asked to write a quadratic equation based on the answer given. I have already done it, but someone is telling me I am wrong, and when I did it again, I came up with a completely different answer! So confused. Anyway, the problem is:
3i +sqrt+%28+2+%29+ , -3i +sqrt+%28+2+%29+
Here are my original steps:
x-3i +sqrt+%28+2+%29+ =0 or x+3i +sqrt+%28+2+%29+ =0
(x-3i +sqrt+%28+2+%29+) (x+3i +sqrt+%28+2+%29+) =0
+x%5E2+ -9+i%5E2++2=0
+x%5E2+ +9+2=0
+x%5E2+ +11=0
The second time around I multiplied 9 and 2, coming up with +x%5E2+ +18=0
I am kind of confused on what to do with the square root value. Do you add it to the equation, as I did the first time, or do you multiply, as I did the second time?
Thank you.
Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
You got the right answer the second time...
(x-3isqrt%282%29%29)(x+3isqrt%282%29%29)=0
x%5E2%2B%28-9i%5E2%29%282%29=0
highlight%28x%5E2%2B18=0%29
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Good Luck,
tutor_paul@yahoo.com