Question

if x+1\x =11 find the value of x^2+1\x^2​

Answer 1

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$If \ \left( x+ \displaystyle{\frac{1}{x}} \right) =11\ find\ the\ \ value\ of\\ \left( x^2 + \displaystyle{\frac{1}{x^2}}\right)$

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$\boxed{\red{\bold{Given:}}}$

$\purple{\implies \left( x+ \displaystyle{\frac{1}{x}} \right)= 11}$

$\boxed{\red{\bold{Squaring\ both \ sides:}}}$

$\green{\implies \left( x+ \displaystyle{\frac{1}{x}} \right)^2 = (11)^2}$

$\green{\implies}x^2 + \displaystyle{\frac{1}{x^2}} + 2 × x × \frac{1}{x} = 121$

$\blue{(\because By \ identity \ (a+b)^2 = a^2 + b^2 + 2ab)}$

$\green{\implies}x^2 + \displaystyle{\frac{1}{x^2}} + 2 = 121$

$\green{\implies}x^2 + \displaystyle{\frac{1}{x^2}} = 121 - 2$

$\green{\boxed{\implies{\boxed{x^2 + \displaystyle{\frac{1}{x^2}} = 119}}}}$

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