A) \[3,4\]
B) \[3,-3\]
C) \[3,4,-3,-4\]
D) \[~-3,-3\]
Correct Answer: C
Solution :
We have, \[{{x}^{2}}+{{y}^{2}}=25,xy=12\] \[\Rightarrow \] \[{{x}^{2}}+{{\left( \frac{12}{x} \right)}^{2}}=25\] \[\Rightarrow \] \[{{x}^{4}}-25{{x}^{2}}+144=0\] \[\Rightarrow \]\[{{x}^{4}}-16{{x}^{2}}-9{{x}^{2}}+144=0\] \[\Rightarrow \] \[({{x}^{2}}-16)({{x}^{2}}-9)=0\] \[\Rightarrow \]\[x=\pm \,4\]and \[x=\pm \,3\] Alternate Solution: We have, \[{{x}^{2}}+{{y}^{2}}=25\]and \[xy=12\] Now,\[{{x}^{2}}+{{y}^{2}}+2xy=25+2(12)\] \[\Rightarrow \] \[{{(x+y)}^{2}}=49\] \[\Rightarrow \] \[x+y=\pm \,7\] ?(i) and \[{{x}^{2}}+{{y}^{2}}-2xy=25-2(12)\] \[\Rightarrow \] \[{{(x-y)}^{2}}=1\] \[\Rightarrow \] \[x-y=\pm \,1\] On solving Eqs. (i) and (ii), we get \[x=\pm \,4\]and \[x=\pm \,3\]
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