Math, asked by Anonymous, 1 month ago


\huge\fcolorbox{cyan}{black}{\red{QUESTION}}
IF(x -  \frac{1}{x} ) ^{2}  = 64 \: THEN  \: FIND
THE  \:  \:  VALUE  \:  \: OF \:  \:  \:  {x}^{2 }  +  \frac{1}{ {x}^{2}  }

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Answers

Answered by devanshu1234321
0

EXPLANATION:-

\sf\((x-\frac{1}{x})^2=64\\\\x-\frac{1}{x}=\sqrt{64}          \;\;\;\;\;\;\;\;\;\;[Transpose\;Square\;sign\;to\;LHS]\\\\x-\frac{1}{x}=8\\\\Now\;let's square\;on\;both;sides\\\\(x-\frac{1}{x})^2=8^2\\\\x^2+\frac{1}{x^2}-2=64 \;\;\;\;\;[USING\;(a-b)^2=a^2-b^2=2ab]\\\\x^2+\frac{1}{x}^2=64+2\;\;\;\;[TRANSPOSE -2 \;TO\' LHS]\\\\x^2+\frac{1}{x}^2=66

Hence ,the value of x²+1/x² is 66

Answered by vshouryansh6
0

Answer:

the value is 66

hope it helps

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