If $K +\frac{1}{K} + 2 = 0 $ and K < 0, then what is the value of $ K^{11}+\frac{1}{K^4}$ ?
Answer & explanation
Correct answer: option 1
$K +\frac{1}{K} + 2 = 0 $
$K +\frac{1}{K} = -2 $
Put k = -1
$ -1^{11}+\frac{1}{-1^4}$ = 0
If $K +\frac{1}{K} + 2 = 0 $ and K < 0, then what is the value of $ K^{11}+\frac{1}{K^4}$ ?
Correct answer: option 1
$K +\frac{1}{K} + 2 = 0 $
$K +\frac{1}{K} = -2 $
Put k = -1
$ -1^{11}+\frac{1}{-1^4}$ = 0