1/K+4 + K/K-4 = -8 / K^2-16?

Question

Solve for K and Check

1/K+4 + K/K-4 = -8 / K^2-16

Answer 1

1/K+4 + K/K-4 = -8 / K^2-16

(K-4+K+4)/(K^2-16) =-8/K^2-16

2K =-8

K = -8/2

K = -4

Answer 2

Presentation is INCORRECT I would suspect and should be :-



1 / ( k + 4 ) + k / ( k - 4 ) = - 8 / ( k ² - 16 )



1 / ( k + 4 ) + k / ( k - 4 ) = - 8 / [ ( k - 4 ) ( k + 4 ) ]



( k - 4 ) + k ( k + 4 ) = - 8



k - 4 + k ² + 4k = - 8



k ² + 5k + 4 = 0



( k + 4 ) ( k + 1 ) = 0



k = - 4 , k = - 1



Accept k = - 1



LHS

---------

1/3 + (1/5) = 8/15



RHS

-------

- 8 / (-15) = 8/15

Answer 3

... 1/(k+4) + k/(k-4) = -8 /(k^2-16) ← k ≠ ± 4 else entire equation is undefined

or 1/(k+4) + k/(k-4) = -8 /[ (k+4)(k-4) ] ← multiply both sides by (k+4)(k-4)

or (k-4) + k(k+4) = -8

or k - 4 + k² + 4k + 8 = 0

or k² + 5k + 4 = 0

or (k + 4)(k + 1) = 0

or k = -4 ← reject as above

or k = -1



Check:

1/(-1+4) + -1/(-1-4) = 1/3 + 1/5 = 8/15 ?=? -8 /((-1)^2-16) = 8/15 ... Okay

Answer 4

1/(K+4)+1/(K-4)=-8/k^2-16

=> ((K-4)+(k+4))/(K+4)(K-4)=-8/(k^2-16)

=> ((K-4)+(k+4))/(k^2-16)=-8/(k^2-16)

=> K-4+K+4=-8 ( since K^2-16 is cancelled at both sides at denominator)

=> 2K=-8 (4 and -4 are also get cancelled out)

=> K=-8/2

=> K=-4

Answer 5

((K-4)/(K²-16)) + (K(K+4)/(K²-16)) = -8/ K²-16

(K-4 +K(K+4)) / (K² -16) = -8/ K²-16



Compare



K-4 +K(K+4) = -8

K² +4K +K - 4 + 8 = 0

K² + 5K + 4 = 0

(K+1)(K+4) =0

K = -1, K = -4



Check it by substituing back the values into the above equation

Answer 6

Misa is correct.