(2k+1)/(18)+(3)/(4k+2)

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Solution for (2k+1)/(18)+(3)/(4k+2) equation: 


D( k )

4*k+2 = 0

4*k+2 = 0

4*k+2 = 0

4*k+2 = 0 // - 2

4*k = -2 // : 4

k = -2/4

k = -1/2

k in (-oo:-1/2) U (-1/2:+oo)

(2*k+1)/18+3/(4*k+2) = 0

((2*k+1)*(4*k+2))/(18*(4*k+2))+(3*18)/(18*(4*k+2)) = 0

(2*k+1)*(4*k+2)+3*18 = 0

8*k^2+8*k+56 = 0

8*k^2+8*k+56 = 0

8*(k^2+k+7) = 0

k^2+k+7 = 0

DELTA = 1^2-(1*4*7)

DELTA = -27

DELTA < 0

8 = 0

8/(18*(4*k+2)) = 0

8/(18*(4*k+2)) = 0 // * 18*(4*k+2)

8 = 0

k belongs to the empty set

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