A) 100
B) 88
C) 96
D) 98
Correct Answer: D
Solution :
\[x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\] \[\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{{{2}^{2}})}}{(\sqrt{3{{)}^{2}}-(\sqrt{2{{)}^{2}}}}}\] \[=\frac{3+2+2\sqrt{3\,}\sqrt{2}}{3-2}=5+2\sqrt{6}\] Similarly, we have \[y=5-2\sqrt{6}\] \[\therefore \] \[x+y=10\] and \[xy=25-24=1\] \[\therefore \]\[{{x}^{2}}+{{y}^{2}}={{(x+y)}^{2}}-2\times y\] \[={{(10)}^{2}}-2\times 1=100-2=98\]
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