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If \( \Large x=3-\sqrt{5} \), then value of \( \Large x^{2}+\frac{16}{x^{2}} \) is equal to

A) 10
B) 24
C) 26
D) 28

Correct Answer:




D) 28

Description for Correct answer:
Given : \( \Large x=3-\sqrt{5} \)

\( \Large \therefore \frac{1}{x}=\frac{1}{3-\sqrt{5}}=\frac{1}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}} \)

= \( \Large \frac{3+\sqrt{5}}{9-5}=\frac{1}{4} \left(3+\sqrt{5}\right) \)

=> \( \Large \frac{4}{x}=3+\sqrt{5} \)

\( \Large \therefore x^{2}+\frac{16}{x^{2}}= \left(3-\sqrt{5}\right)^{2}+ \left(3+\sqrt{5}\right)^{2} \)

= \( \Large \left(9+5-6\sqrt{5}\right)+ \left(9+5+6\sqrt{5}\right) \)

\( \Large = 28 \)

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

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