A) 24
B) 144
C) 128
D) 80
Correct Answer: D
Solution :
\[x+y=12;\,xy=32\] \[\therefore \]\[{{x}^{2}}+{{y}^{2}}={{(x+y)}^{2}}-2xy\] \[={{(12)}^{2}}-2\times 32\] \[=144-64=80\]You need to login to perform this action.
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