Question
If x = \(\frac{\sqrt{3}+1}{\sqrt{3}-1} and y = \frac{\sqrt{3}-1}{\sqrt{3}+1} , then the value of (x^{2}+y^{2}) is:\)
Answer: Option C
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\(x =\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)}{(3-1)}^{2} = \frac{3+1+2\sqrt{3}}{2} = 2+\sqrt{3}.\)
\(y =\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}\times\frac{(\sqrt{3}-1)}{(\sqrt{3}-1)} = \frac{(\sqrt{3}-1)}{(3-1)}^{2} = \frac{3+1-2\sqrt{3}}{2} = 2-\sqrt{3}\)
\(\therefore a^{b}+a^{b} = (2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}\)
= 2(4 + 3)
= 14
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