If x+frac{1}{x}=2, then find the value of x^{ | vedclass

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(A) Given,$x+\frac{1}{x}=2.$Squaring both sides,we get:$\left(x+\frac{1}{x}\right)^{2} = (2)^{2}$$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\righ

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$2$

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(A) Given,$x+\frac{1}{x}=2.$Squaring both sides,we get:$\left(x+\frac{1}{x}\right)^{2} = (2)^{2}$$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 4$$x^{2} + \frac{1}{x^{2}} + 2 = 4$$x^{2} + \frac{1}{x^{2}} = 2$Now,squaring both sides again:$\left(x^{2} + \frac{1}{x^{2}}\right)^{2} = (2)^{2}$$x^{4} + \frac{1}{x^{4}} + 2(x^{2})\left(\frac{1}{x^{2}}\right) = 4$$x^{4} + \frac{1}{x^{4}} + 2 = 4$$x^{4} + \frac{1}{x^{4}} = 2.$

If $x+\frac{1}{x}=2,$ then find the value of $x^{4}+\frac{1}{x^{4}}.$

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