SOLUTION: How does one evaluate (what is method for) following question...? "If k is a negative number, which of following equations will have nonreal complex solutions: a) x^2 = 4k, b) x

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Question 862078: How does one evaluate (what is method for) following question...?
"If k is a negative number, which of following equations will have nonreal complex solutions: a) x^2 = 4k, b) x^2=-4k, c) (x+2)^2=-k, d) x^2+k=0 ."

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If x is a real number, x%5E2 and %28x%2B2%29%5E2 will be non-negative numbers, so for k%3C0 ,
a) x%5E2+=+4k%3C0 cannot have a real solution.
It has non-real complex solutions.

For the other choices, k%3C0 makes the squares equal to a positive number
b) x%5E2=-4k%3E0
c) %28x%2B2%29%5E2=-k%3E0
d) x%5E2%2Bk=0<--->x%5E2=-k%3E0
Each one of those equations has two real solutions.

NOTE:
If there was not an obvious square, you would have to make one appear by "completing the square".
Sometimes it is easy as in
x%5E2%2B2x=k-1<-->x%5E2%2B2x%2B1=k<-->%28x%2B1%29%5E2=k.
In other cases, you may want to use the work of ancient mathematicians that figured out a general formula to complete those squares.
It turns out that for an equation that can be written as
ax%5E2%2Bbx%2Bc=0
with any coefficients
a%3C%3E0 , b, and c ,
the square will have the same sign as
b%5E2-4ac ,
so if b%5E2-4ac%3C0 ,
the equation ax%5E2%2Bbx%2Bc=0
has non-real complex solutions.