If \( \Large x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\ and\ y=\frac{\sqrt{3}-1}{\sqrt{3}+1} \), the value of \( \Large x^{2}+y^{2} \) is:
Correct Answer: Description for Correct answer:
\( \Large x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\ and\ y=\frac{\sqrt{3}-1}{\sqrt{3}+1} \)
=> Therefore, \( \Large x = \frac{1}{y} \)
\( \Large x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
= \( \Large \frac{ \left(\sqrt{3}+1\right)^{2} }{3-1} \)
= \( \Large \frac{3+1+2\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{2} \)
= \( \Large \left(2 + \sqrt{3}\right) \)
\( \Large x^{2}= \left(2+\sqrt{3}\right)^{2} = 4+3+4\sqrt{3} \)
= \( \Large 7 + 4\sqrt{3} \)
\( \Large y^{2}=\frac{1}{7+4\sqrt{3}} \times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} \)
\( \Large y^{2} = \frac{7-4\sqrt{3}}{49-48} = \frac{7-4\sqrt{3}}{1} = 7 - 4\sqrt{3} \)
Therefore, \( \Large x^{2}+y^{2}=7+4\sqrt{3}+7-4\sqrt{3} = 14 \)
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