If \( x^{2}+\frac{1}{x^{2}}=79 \), find the value of \( x+\frac{1}{x} \).

Given:

\( x^{2}+\frac{1}{x^{2}}=79 \)

To do:

We have to find the value of \( x+\frac{1}{x} \).

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,

$(x+\frac{1}{x})^{2}=x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}$

$\Rightarrow (x+\frac{1}{x})^{2}=79+2$

$\Rightarrow (x+\frac{1}{x})^{2}=81$

$\Rightarrow x+\frac{1}{x}=\sqrt{81}$

$\Rightarrow x+\frac{1}{x}=\pm 9$

Hence, $x+\frac{1}{x}=\pm 9$. 

Tutorialspoint

Updated on: 10-Oct-2022

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