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Question: Solving cubic equation

Question:Solving cubic equation

Syeda 30
solve rootfinding roundoff numeric
March 18 2014
0

Hi all,

By solving cubic equation in maple (version 17), I got

restart

``

-0.363700352e-2*y^3-.4041941000*y^2+3.397775673*y-2.377540486 = 0

-0.363700352e-2*y^3-.4041941000*y^2+3.397775673*y-2.377540486 = 0

 (1)

"(->)"

[[y = .7709248124], [y = 7.123944371], [y = -119.0286907]]

 (2)

``

Now I want to find these roots through the formula.

 

I solve it generally in Maple.. 

 

``# Suppose

A*y^3+B*y^2+C*y+E = 0

A*y^3+B*y^2+C*y+E = 0

 (3)

NULL

A := -0.363700352e-2:

B := -.4041941000:

C := 3.397775673:

E := -2.377540486:

``

A*y^3+B*y^2+C*y+E = 0

 

A*y^3+B*y^2+C*y+E = 0

 (4)

``

y1 := (1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A-(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A

-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717

 (5)

"(=)"

-119.0286907-0.1e-8*I

 (6)

y2 := y = -(1/12)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(1/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A+(1/2*I)*sqrt(3)*((1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)))

y = 22.91263477*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+18.37098733/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717+((1/2)*I)*3^(1/2)*(-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3))

 (7)

"(=)"

y = .770924807+0.1772050808e-7*I

 (8)

y3 := y = -(1/12)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(1/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A-(1/2*I)*sqrt(3)*((1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)))

y = 22.91263477*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+18.37098733/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717-((1/2)*I)*3^(1/2)*(-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3))

 (9)

"(=)"

y = 7.123944373-0.1692050808e-7*I

 (10)

``


y1, y2, y3 formulas are computed by Maple by solving it for general formula.
But, now I got answers in real and imaginery parts, i.e

 

y1 = -119.0286907-1.*10^(-9)*I

y2 = .770924807+1.772050808*10^(-8)*I

y3 = 7.123944373-1.692050808*10^(-8)*I

 

Why, is it so?

 

 

I want answers in simple forum directly only by using these formulas. As i have to show the proof!

Thanks in advance

 

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