y^2-4/y+3

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Solution for y^2-4/y+3 equation:


D( y )

y = 0

y = 0

y = 0

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

y^2-(4/y)+3 = 0

y^2-4*y^-1+3 = 0

1*y^2-4*y^-1+3*y^0 = 0

(1*y^3+3*y^1-4*y^0)/(y^1) = 0 // * y^2

y^1*(1*y^3+3*y^1-4*y^0) = 0

y^1

y^3+3*y-4 = 0

{ 1, -1, 2, -2, 4, -4 }

1

y = 1

y^3+3*y-4 = 0

1

y-1

y^2+y+4

y^3+3*y-4

y-1

y^2-y^3

y^2+3*y-4

y-y^2

4*y-4

4-4*y

0

y^2+y+4 = 0

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

DELTA = -15

DELTA < 0

y in { 1}

y = 1

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