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

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Solution for 4k+8/(4k-1)(k+2) equation: 


D( k )

4*k-1 = 0

4*k-1 = 0

4*k-1 = 0

4*k-1 = 0 // + 1

4*k = 1 // : 4

k = 1/4

k in (-oo:1/4) U (1/4:+oo)

(8/(4*k-1))*(k+2)+4*k = 0

(8*(k+2))/(4*k-1)+4*k = 0

(8*(k+2))/(4*k-1)+(4*k*(4*k-1))/(4*k-1) = 0

8*(k+2)+4*k*(4*k-1) = 0

16*k^2+4*k+16 = 0

16*k^2+4*k+16 = 0

4*(4*k^2+k+4) = 0

4*k^2+k+4 = 0

DELTA = 1^2-(4*4*4)

DELTA = -63

DELTA < 0

4 = 0

4/(4*k-1) = 0

4/(4*k-1) = 0 // * 4*k-1

4 = 0

k belongs to the empty set

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