(k^3+4k^2-46k-5)/(k-5)

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Solution for (k^3+4k^2-46k-5)/(k-5) equation:


D( k )

k-5 = 0

k-5 = 0

k-5 = 0

k-5 = 0 // + 5

k = 5

k in (-oo:5) U (5:+oo)

t_2 = 0

(k^3+4*k^2-46*k-5)/(k-5)+t_2 = 0

(k^3+4*k^2-46*k-5)/(k-5)+(t_2*(k-5))/(k-5) = 0

t_2*(k-5)+k^3+4*k^2-46*k-5 = 0

k^3+4*k^2-46*k+k*t_2-5*t_2-5 = 0

k^3+4*k^2-46*k+k*t_2-5*t_2-5 = 0

k^3+4*k^2-46*k+k*t_2-5*t_2-5

k*(k^2-46)+4*k^2+k*t_2-5*t_2-5

k*(4*k+t_2)+k*(k^2-46)-5*t_2-5

k*(4*k+t_2)-5*(t_2+1)+k*(k^2-46)

k*(k^2+4*k+t_2-46)-5*(t_2+1)

-4*(k^2+4*k+t_2-46)

(-4*(k^2+4*k+t_2-46))/(k-5) = 0

(-4*(k^2+4*k+t_2-46))/(k-5) = 0 // * k-5

-4*(k^2+4*k+t_2-46) = 0

k^2+4*k+t_2-46 = 0

DELTA = 4^2-(1*4*(t_2-46))

DELTA = 16-4*(t_2-46)

16-4*(t_2-46) = 0

16-4*(t_2-46) = 0

200-4*t_2 = 0

200-4*t_2 = 0

200-4*t_2 = 0 // - 200

-4*t_2 = -200 // : -4

t_2 = -200/(-4)

t_2 = 50

DELTA = 0 <=> t_4 = 50

k = -4/(1*2) i t_2 = 50

k = -2 i t_2 = 50

( k = ((16-4*(t_2-46))^(1/2)-4)/(1*2) or k = (-(16-4*(t_2-46))^(1/2)-4)/(1*2) ) i t_2 > 50

( k = ((16-4*(t_2-46))^(1/2)-4)/2 or k = (-((16-4*(t_2-46))^(1/2)+4))/2 ) i t_2 > 50

t_2-50 > 0

t_2-50 > 0 // + 50

t_2 > 50

k in { -2, ((16-4*(0-46))^(1/2)-4)/2, (-((16-4*(0-46))^(1/2)+4))/2 }

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