Question

14. If x²+1/x²=27 find the value of x-1/x​

Answer 1

EXPLANATION.

$\sf : \implies \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 27 \\ \\ \sf : \implies \: { \underline{then \: value \: of \: x - \frac{1}{x} }}$

$\sf : \implies \: (x - \dfrac{1}{x}) {}^{2} = {x}^{2} - 2 \times \: x \: \times \dfrac{1}{x} + \dfrac{1}{ {x}^{2} } \\ \\ \sf : \implies \: (x - \frac{1}{x} ) {}^{2} = {x}^{2} - 2 + \frac{1}{ {x}^{2} } \\ \\ \sf : \implies \: (x - \frac{1}{x}){}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ \sf : \implies \: (x - \frac{1}{x} ) {}^{2} = 27 - 2 \\ \\\sf : \implies(x - \frac{1}{x}) {}^{2} = 25$

$\sf : \implies \: (x - \dfrac{1}{x}) = \sqrt{25} \\ \\\sf : \implies \: (x - \frac{1}{x}) {}^{2} = \pm \: 5 \\ \\ \sf : \implies \: \orange{{ \underline{the \: value \: of \: (x - \frac{1}{x}) = \: \: \pm \: 5}}}$

Answer 2

Knowledge Required :

$\bullet\ \; \tt \bigg(x-\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}-2.\cnacel{x}.\dfrac{1}{\cancel{x}}\\\\\bullet\ \; \tt \bigg(x-\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}-2$

Solution :

$\tt x^2+\dfrac{1}{x^2}=27\\\\\to \tt x^2+\dfrac{1}{x^2}=25+2\\\\\to \tt x^2+\dfrac{1}{x^2}-2=25\\\\\to \tt x^2+\dfrac{1}{x^2}-2.x.\dfrac{1}{x}=25\\\\\to \tt \bigg(x-\dfrac{1}{x}\bigg)^2=25\\\\\to \tt x-\dfrac{1}{x}=\pm \sqrt{25}\\\\\leadsto \tt \pink{x-\dfrac{1}{x}=\pm 5}\ \; \bigstar$