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

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(B) Given that $x+\frac{1}{x}=2.$To find the value of $x^{2}+\frac{1}{x^{2}},$ we square both sides of the given equation:$\left(x+\frac{1}{x}\right)^

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$2$

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(B) Given that $x+\frac{1}{x}=2.$To find the value of $x^{2}+\frac{1}{x^{2}},$ we square both sides of the given equation:$\left(x+\frac{1}{x}\right)^{2} = 2^{2}$Using the algebraic identity $(a+b)^{2} = a^{2}+b^{2}+2ab,$$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 4$$x^{2} + \frac{1}{x^{2}} + 2 = 4$$x^{2} + \frac{1}{x^{2}} = 4 - 2$$x^{2} + \frac{1}{x^{2}} = 2$

If $x+\frac{1}{x}=2,$ then the value of $x^{2}+\frac{1}{x^{2}}$ will be

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