Question 889047
(x+1)^2=-k will have real solutions since k < 0 leads to -k > 0


So choice A is out.


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2x^2+5k=0


2x^2=-5k


will have real solutions since -5k is also positive (same reasoning as A)


choice B is out


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There's a typo in choice C


You wrote (x-5)^2=k=0


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k-3x^2=0


3x^2 = k


this leads to two imaginary solutions since k < 0