Math, asked by LuvBazzar, 7 months ago

If x=3-√8, then (x-1/x)^2​

Answers

Answered by Asterinn
5

Given :

 \rm x = 3 -  \sqrt{8}

To find :

 \rm { \bigg(x +  \dfrac{1}{x} \bigg) }^{2}

Solution :

\rm x = 3 -  \sqrt{8}  \\  \\ \rm  \rightarrow \frac{1}{x}  =  \frac{1}{3 -  \sqrt{8}} \\  \\ \rm  \rightarrow \frac{1}{x}  =  \frac{1}{3 -  \sqrt{8}}  \times \frac{3  +   \sqrt{8}}{3  +   \sqrt{8}} \\  \\ \rm  \rightarrow \frac{1}{x}  =   \dfrac{3  +   \sqrt{8}}{9   -   8} \\  \\ \rm  \rightarrow \frac{1}{x}  =   \dfrac{3  +   \sqrt{8}}{1} \\  \\ \rm  \rightarrow \frac{1}{x}  =3  +   \sqrt{8}

\rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  =  {x}^{2}  + \dfrac{1}{{x}^{2}} +( 2 \times x \times  \dfrac{1}{x}) \\  \\ \rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  =  {x}^{2}  + \dfrac{1}{{x}^{2}} +2  \\  \\ \\  \\ \rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  =  {(3 -  \sqrt{8} )}^{2}  + {(3  +  \sqrt{8} )}^{2} +2  \\  \\ \rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  = 9 + 8 - 6 \sqrt{8}  + 9 + 8  +  6 \sqrt{8} +2  \\  \\ \rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  = 9 + 8  + 9 + 8 +2 \\  \\ \rm \rightarrow{ \bigg(x +  \dfrac{1}{x} \bigg) }^{2}  = 36

Answer :

36

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