If x=sqrt{3},then the value of x^{4}+2+frac{1 | vedclass

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(D) The expression is $x^{4}+2+\frac{1}{x^{4}}$.This can be rewritten as $(x^{2})^{2} + 2(x^{2})(\frac{1}{x^{2}}) + (\frac{1}{x^{2}})^{2}$.This is in

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$\frac{100}{9}$

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(D) The expression is $x^{4}+2+\frac{1}{x^{4}}$.This can be rewritten as $(x^{2})^{2} + 2(x^{2})(\frac{1}{x^{2}}) + (\frac{1}{x^{2}})^{2}$.This is in the form of the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$,where $a = x^{2}$ and $b = \frac{1}{x^{2}}$.So,the expression becomes $(x^{2} + \frac{1}{x^{2}})^{2}$.Given $x = \sqrt{3}$,then $x^{2} = 3$ and $\frac{1}{x^{2}} = \frac{1}{3}$.Substituting these values,we get $(3 + \frac{1}{3})^{2}$.$= (\frac{9+1}{3})^{2} = (\frac{10}{3})^{2}$.$= \frac{100}{9}$.

If $x=\sqrt{3}$,then the value of $x^{4}+2+\frac{1}{x^{4}}$ will be

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