Question

If x-$\frac{1}{x}$ =6 Then the value of $x^{2}$ + $\frac{1}{x^{2} }$ is

Answer 1

Question :-

If $\sf x - \dfrac{1}{x} = 6$, then the value of $\sf x^2 + \dfrac{1}{x^2}$ is

Answer :-

We are given,

➩ $\sf x - \dfrac{1}{x} = 6$

Squaring on both the side :-

➩ $\sf \Bigg(x - \dfrac{1}{x}\Bigg)^2 = (6)^2$

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• Identity used - $\sf (a - b)^2 = a^2+b^2-2ab$

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➩ $\sf (x)^2 + \Bigg(\dfrac{1}{x}\Bigg)^2 - 2 \times \cancel x \times \dfrac{1}{\cancel x} = 6$

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➩ $\sf x^2 + \dfrac{1}{x^2} - 2 = 6$

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➩ $\sf x^2 + \dfrac{1}{x^2} = 6 + 2$

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➩ $\boxed{\sf x^2 + \dfrac{1}{x^2} = 8 }$

Answer 2

Answer:

$➩ \sf x - \dfrac{1}{x} = 6x−$

$➩ \sf \Bigg(x -\dfrac{1{x\Bigg)^2=(6)^2(x[tex]</p><p></p><p></p><p></p><p></p><p>⠀</p><p></p><p>⠀</p><p></p><p>[tex] \sf \: Identity \: used - \sf (a - b)^2 = a^2+b^2-2ab(a−b)$

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$➩ \sf (x)^2 + \Bigg(\dfrac{1}{x}\Bigg)^2 - 2 \times \cancel x \times \dfrac{1}{\cancel x} </p><p>$

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$➩ \sf x^2 + \dfrac{1}{x^2}$

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$➩ \sf x^2 + \dfrac{1}{x^2}$

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$➩ \boxed{\sf x^2 + \dfrac{1}{x^2} = 8 }$