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If $ x+\frac{1}{x}=\sqrt{5} $, find the values of $ x^{2}+ \frac{1}{x^{2}} $ and $ x^{4}+\frac{1}{x^{4}} $.
Given:
\( x+\frac{1}{x}=\sqrt{5} \)
To do:
We have to find the values of \( x^{2}+\frac{1}{x^{2}} \) and \( x^{4}+\frac{1}{x^{4}} \).
Solution:
We know that,
$(a+b)^2=a^2+b^2+2ab$
$(a-b)^2=a^2+b^2-2ab$
$(a+b)(a-b)=a^2-b^2$
Therefore,
$x+\frac{1}{x}=\sqrt{5}$
Squaring both sides, we get,
$(x+\frac{1}{x})^{2}=(\sqrt{5})^{2}$
$\Rightarrow x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=5$
$\Rightarrow x^{2}+\frac{1}{x^{2}}+2=5$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=5-2=3$
Squaring both sides, we get,
$(x^{2}+\frac{1}{x^{2}})^{2}=(3)^{2}$
$\Rightarrow x^{4}+\frac{1}{x^{4}}+2\times x^2 \times \frac{1}{x^2}=9$
$\Rightarrow x^{4}+\frac{1}{x^{4}}=9-2=7$
The values of \( x^{2}+\frac{1}{x^{2}} \) and \( x^{4}+\frac{1}{x^{4}} \) are $3$ and $7$ respectively.
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