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If \( \Large x = 7-4\sqrt{3} \), then the value of \( \Large x+\frac{1}{x} \)
A) \( \Large 3\sqrt{3} \)
B) \( \Large 8\sqrt{3} \)
C) \( \Large 14+8\sqrt{3} \)
D) 14
Correct answer:
D) 14
Description for Correct answer:
\( \Large x = 7 - 4\sqrt{3} \)
\( \Large \frac{1}{x} => \frac{1}{7-4\sqrt{3}} \)
By ratonalisation,
\( \Large \frac{1}{x}=\frac{1}{7-4\sqrt{3}} \times \frac{7+4\sqrt{3}}{7+4\sqrt{3}} \)
= \( \Large \frac{7+4\sqrt{3}}{49-48}= 7 + 4\sqrt{3} \)
Therefore, \( \Large x+\frac{1}{x} = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} \)
= 14
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