SOLUTION: I really need help to solve the following -k(-2-3) - (-2)(-k-5) = -2-(-2k+4) + 3 times the square root of -27

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Question 62240This question is from textbook alegbra concepts and applications
: I really need help to solve the following
-k(-2-3) - (-2)(-k-5) = -2-(-2k+4) + 3 times the square root of -27 This question is from textbook alegbra concepts and applications

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
-k(-2-3) - (-2)(-k-5) = -2-(-2k+4) + 3 times the square root of -27
:
-k(-2-3) - (-2)(-k-5) = -2 - 1(-2k+4) + 3*SqRt(-27)
:
k(-2-3) - (-2)(-k-5) = -2 - 1(-2k+4) + 3*SqRt(3*3*3*-1)
:
-k(-5) + 2(-k-5) = -2 + 2k - 4 + 3*3i*SqRt(3)
:
+5k - 2k - 5 = - 2 + 2k - 4 + 9i*SqRt(3)
:
5k - 2k - 2k = - 6 + 5 + 9i*SqRt(3)
:
k = -1 + 9i*SqRt(3)
:
The main thing is watch the signs and change them if there is a neg outside the brackets. I guess you know that i = SqRt(-1)