Question

50 POINTS!!!
algebraic identities



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Answer 1

Answer:

this is your solution for the following information

Answer 2

The value of $x-\frac{1}{x} =\pm 5$

Step-by-step explanation:

Given $x^{2} +\frac{1}{x^{2} } =27$           ..............Equ(1)

   we have to find the value of $x-\frac{1}{x}$

We know $(a+b)^{2} =a^{2} +b^{2} +2ab$

so $a^{2} +b^{2} =(a+b)^{2} -2ab$

Equ(1) can be written as

    $(x-\frac{1}{x}) ^{2} +2\times x \times(\frac{1}{x} )=27$

or $(x-\frac{1}{x}) ^{2} =27-2$

 or $(x-\frac{1}{x} )^{2} =25$

or $x-\frac{1}{x} =\sqrt{25}$

or $x-\frac{1}{x} =\pm 5$

The value of $x-\frac{1}{x} =\pm 5$