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Mathematics

If x2+1x2=27x^2 + \dfrac{1}{x^2} = 27, find the value of x1xx - \dfrac{1}{x}.

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Answer

By formula,

(x1x)2=x2+1x22\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2

Substituting values in above equation, we get :

(x1x)2=272(x1x)2=25(x1x)=25(x1x)=±5.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 27 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 25 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{25} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 5.

Hence, (x1x)=±5.\Big(x - \dfrac{1}{x}\Big) = \pm 5.

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