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:


A) 14

B) 13

C) 15

D) 10

Correct Answer:
A) 14

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, \( \Large x^{2}+y^{2}=7+4\sqrt{3}+7-4\sqrt{3} = 14 \)

Part of solved Elementary Mathematics questions and answers : >> Elementary Mathematics








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