Question

if x-1/x =16, then, what is x2 + 1/x2

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x-\dfrac{1}{x}=16}$

$\tt{\implies\,\left(x-\dfrac{1}{x}\right)^2=(16)^2}$

$\tt{\implies\,x^2+\dfrac{1}{x^2}-2\cdot\,x\cdot\dfrac{1}{x}=256}$

$\tt{\implies\,x^2+\dfrac{1}{x^2}-2=256}$

$\tt{\implies\,x^2+\dfrac{1}{x^2}=256+2}$

$\tt{\implies\,x^2+\dfrac{1}{x^2}=258}$

Answer 2

Answer:

258

Step-by-step explanation:

We have,

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

square on both sides,

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

$x^2 +\bigg( \dfrac{1}{x} \bigg)^2 - 2·x· \dfrac{1}{x} = 256$

$x^2 +\bigg( \dfrac{1}{x} \bigg)^2 - 2·\cancel{x}· \dfrac{1}{\cancel{x}} = 256$

$x^2 + \dfrac{1}{x^2} -2 = 256$

$x^2 + \dfrac{1}{x^2} = 256+2$

$x^2 + \dfrac{1}{x^2} = 258$