Ifthen find the value of ( i )( ii )

1.	Ifthen find the value of ( i )( ii )
Answer» GIVEN:- $=> (x - \frac{1}{x}) = 9$ To FIND:- $=> ({x}^{2} + \frac{1}{ {x}^{2} })$ $=> ({x}^{4} + \frac{1}{ {x}^{4} })$ Solution:- By squaring both the sides in the equation (x - 1/x)=9, we get:- $= > (x - \frac{1}{x} ) ^{2} = (9)^{2}$ We know that:- =>(a-b)² = a² + b² - 2ab $= > (x) ^{2} + ({ \frac{1}{x} )}^{2} - 2 = 81$ $= > {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 81$ $= > {x}^{2} + \frac{1}{ {x}^{2}} = 81 + 2$ $= > {x}^{2} + \frac{1}{ {x}^{2} } = <klux>83</klux>$ Hence, we have got the VALUE of (x²+1/x²) as 83. _____________________________________ Now,by squaring both the sides in the equation (x² + 1/x²)=83, we get:- $= > ( {x}^{2} + \frac{1}{ {x}^{2} } ) ^{2} = (83)^{2}$ We know that:- =>(a+b)² = a² + b² + 2ab $= >({ {x}^{2} })^{2} + (\frac{1}{ {x}^{2}})^{2} + 2= 6889$ $= > {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 6889$ $= > {x}^{4} + \frac{1}{ {x}^{4} } = 6889 - 2$ $= > {x}^{4} + \frac{1}{ {x}^{4} } = 6887$ Thus:- •Value of (x²+1/x²) is 83. •Value of (x⁴+1/x⁴) is 6887.

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