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If $ x^{2}+\frac{1}{x^{2}}=62, $ find the value of
- $ x+\frac{1}{x} $
- $ x-\frac{1}{x} $
Given:
\( x^{2}+\frac{1}{x^{2}}=62 \)
To do:
We have to find the value of
(a) \( x+\frac{1}{x} \)
(b) \( x-\frac{1}{x} \)
Solution:
(a) $x^{2}+\frac{1}{x^{2}}=62$
Adding 2 on both sides, we get,
$x^{2}+\frac{1}{x^{2}}+2=62+2$
$x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=64$
$(x+\frac{1}{x})^2=(8)^2$
This implies,
$x+\frac{1}{x}=8$.
(b) $x^{2}+\frac{1}{x^{2}}=62$
Subtracting 2 on both sides, we get,
$x^{2}+\frac{1}{x^{2}}-2=62-2$
$x^{2}+\frac{1}{x^{2}}-2\times x \times \frac{1}{x}=60$
$(x-\frac{1}{x})^2=(\sqrt{15\times4})^2$
$(x-\frac{1}{x})^2=(2\sqrt{15})^2$
This implies,
$x-\frac{1}{x}=2\sqrt{15}$.
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