2/p^2=1/p^2-4/p

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Solution for 2/p^2=1/p^2-4/p equation: 


D( p )

p = 0

p^2 = 0

p = 0

p = 0

p^2 = 0

p^2 = 0

1*p^2 = 0 // : 1

p^2 = 0

p = 0

p in (-oo:0) U (0:+oo)

2/(p^2) = 1/(p^2)-(4/p) // - 1/(p^2)-(4/p)

4/p+2/(p^2)-(1/(p^2)) = 0

4/p+2/(p^2)-p^-2 = 0

4*p^-1+p^-2 = 0

t_1 = p^-1

1*t_1^2+4*t_1^1 = 0

t_1^2+4*t_1 = 0

DELTA = 4^2-(0*1*4)

DELTA = 16

DELTA > 0

t_1 = (16^(1/2)-4)/(1*2) or t_1 = (-16^(1/2)-4)/(1*2)

t_1 = 0 or t_1 = -4

t_1 = -4

p^-1+4 = 0

1*p^-1 = -4 // : 1

p^-1 = -4

-1 < 0

1/(p^1) = -4 // * p^1

1 = -4*p^1 // : -4

-1/4 = p^1

p = -1/4

t_1 = 0

p^-1+0 = 0

p^-1 = 0

1*p^-1 = 0 // : 1

p^-1 = 0

p należy do O

p = -1/4

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