If \( \Large x = \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}} \), then \( \Large x^{2}-x-1 \) is equal to
Correct Answer: Description for Correct answer:
Here, x = \( \Large \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}} \)
On rationalising the terms given root, we get
x = \( \Large \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}-1} } = \frac{\sqrt{5}+1}{2} \)
Now, substituting the value of x in
\( \Large x^{2} - x - 1. \)
Therefore, \( \Large x^{2} -x - 1 = \left(\frac{\sqrt{5}+1}{2}\right)^{2} - \left(\frac{\sqrt{5}+1}{2}\right) - 1 \)
= \( \Large \frac{5+1+2\sqrt{5}}{4} - \frac{\sqrt{5}+1}{2} - 1 \)
= \( \Large \frac{6 + 2 \sqrt{5} - 2 \sqrt{5} - 2 - 4}{4} = 0 \)
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