A) \[{{x}^{2}}=5\]
B) \[{{y}^{2}}=16\]
C) \[{{z}^{2}}=-5\]
D) \[{{a}^{2}}=6\]
Correct Answer: B
Solution :
(b): \[{{y}^{2}}=16,y=\pm \sqrt{16}\] \[\therefore y=\pm 4\]
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