Question

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Answer 1

Given :-

${x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

To find :-

$x - \dfrac{1}{x}$

Identity to use :-

(a-b)² = a² - 2ab + b²

substituting the values,

$(x - \dfrac{1}{x})^{2} = ({x})^{2} - 2(x)( \dfrac{1}{x}) + (\dfrac{1}{x})^{2}$

$(x - \dfrac{1}{x})^{2} = {x}^{2} - 2( \not x)( \dfrac{1}{ \not x}) + \dfrac{1}{ {x}^{2} }$

$(x - \dfrac{1}{x})^{2} = {x}^{2} - 2 + \dfrac{1}{ {x}^{2} }$

By rearranging,

$(x - \dfrac{1}{x})^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$

By substituting the value for x²+1/x²,

$(x - \dfrac{1}{x})^{2} = 18 - 2$

$(x - \dfrac{1}{x})^{2} = 16$

$(x - \dfrac{1}{x}) = √16$

$(x - \dfrac{1}{x}) = 4$

Some more identities :

(a+b)² = a² + 2ab + b²

a²-b² = (a-b)(a+b)

(x+a)(x+b) = x² + x(a+b) + ab

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca