(6x^3-9x^2+13x-5)/(2x-1)

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Solution for (6x^3-9x^2+13x-5)/(2x-1) equation:


D( x )

2*x-1 = 0

2*x-1 = 0

2*x-1 = 0

2*x-1 = 0 // + 1

2*x = 1 // : 2

x = 1/2

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

(6*x^3-(9*x^2)+13*x-5)/(2*x-1) = 0

(6*x^3-9*x^2+13*x-5)/(2*x-1) = 0

6*x^3-9*x^2+13*x-5 = 0

6*x^3-9*x^2+13*x-5 = 0

{ 1, -1, 5, -5 }

1

x = 1

6*x^3-9*x^2+13*x-5 = 5

1

-1

x = -1

6*x^3-9*x^2+13*x-5 = -33

-1

5

x = 5

6*x^3-9*x^2+13*x-5 = 585

5

-5

x = -5

6*x^3-9*x^2+13*x-5 = -1045

-5

{ 1/2, -1/2, 1/3, -1/3, 1/6, -1/6, -1/2, 1/2, -1/3, 1/3, -1/6, 1/6, 5/2, -5/2, 5/3, -5/3, 5/6, -5/6, -5/2, 5/2, -5/3, 5/3, -5/6, 5/6 }

1/2

x

1/2

6*x^3-9*x^2+13*x-5 = 0

1/2

x-1/2

6*x^2-6*x+10

6*x^3-9*x^2+13*x-5

x-1/2

3*x^2-6*x^3

13*x-6*x^2-5

6*x^2-3*x

10*x-5

5-10*x

0

6*x^2-6*x+10 = 0

DELTA = (-6)^2-(4*6*10)

DELTA = -204

DELTA < 0

x in { 1/2}

x-1/2 = 0

(x-1/2)/(2*x-1) = 0

x-1/2 = 0 // + 1/2

x = 1/2

x in { 1/2}

x belongs to the empty set

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