Question

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

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Answer 1

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

Answer 2

Answer:

the value is 66

hope it helps