Solve the following:

If $x^{2}+\frac{1}{x^{2}}=3,$ find a) $ x-\frac{1}{x}$b) $x+\frac{1}{x} $

Solution:

a)

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

= $x^{2} + \frac{1}{x}^{2} - 2 $

= $3 - 2$

= 1

So ,$(x - \frac{1}{x} ) $= 1 or -1

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

= $x^{2} + \frac{1}{x}^{2} + 2 $

= 3 + 2 = 5

So ($x + \frac{1}{x}$ ) = $√5$ or -√5

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Updated on: 10-Oct-2022

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