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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 ($x + \frac{1}{x}$ ) = or -√5
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