If $\frac{1}{(x-2)} + x = 8 $, then what is the value of $\frac{1}{(x-2)^2}+(x - 2)^2 $ ?
Answer & explanation
Correct answer: option 4
If $K+\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 – 2
If $\frac{1}{(x-2)} + x = 8 $,
then what is the value of $\frac{1}{(x-2)^2}+(x - 2)^2 $ ?
Subtract 2 from both the sides,
$\frac{1}{(x-2)} + (x-2) = 8-2 $,
$\frac{1}{(x-2)} + (x-2) = 6 $,
$\frac{1}{(x-2)^2}+(x - 2)^2 $ = 62 – 2 = 34