When x+frac{1}{x}=3,find x^{2}+frac{1}{x^{2}} | vedclass

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(D) Given that $x+\frac{1}{x}=3$.To find the value of $x^{2}+\frac{1}{x^{2}}$,we square the given equation on both sides:$\left(x+\frac{1}{x}\right)^{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given that $x+\frac{1}{x}=3$.To find the value of $x^{2}+\frac{1}{x^{2}}$,we square the given equation on both sides:$\left(x+\frac{1}{x}\right)^{2} = 3^{2}$Using the algebraic identity $(a+b)^{2} = a^{2}+b^{2}+2ab$,we get:$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 9$$x^{2} + \frac{1}{x^{2}} + 2 = 9$Subtracting $2$ from both sides:$x^{2} + \frac{1}{x^{2}} = 9 - 2 = 7$Therefore,the value is $7$.

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

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