Ken Ward's Mathematics Pages

Examples Using Cardano's Method to Solve Cubic Equations

1. 64x³–48x²+12x–1
2.  x³+6x²+11x+6
3. x³+3x²+3x–2
4. x³+2x²+10x–20
5.  x³+x²+10x–3
6. x³+4x²+x–6
7. 1000x³–1254x²–496x+191

64x³–48x²+12x–1

Cubic in normal form: x³–0.75x²+0.1875x–0.015625
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=0 q=0
So the new equation is:
t³+0t +0 [2]
As p and q are zero, then all the roots are real and equal to –a/3 (0.25)
In summary, the roots for
Cubic: 64x³–48x²+12x–1 are:
x1=0.25 (Remainder=0)
x2=0.25 (Remainder=0)
x3=0.25 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

 x³+6x²+11x+6

Cubic in normal form: x³+6x²+11x+6
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=–1 q=0
So the new equation is:
t³–1t +0 [2]
x1=–a/3, x2=–a/3± √(–p)
As p<0, then all the roots are real
In summary, the roots for
x³+6x²+11x+6 are:
x1=–2 (Remainder=0)
x2=–1 (Remainder=0)
x3=–3 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+3x²+3x–2

Cubic in normal form: x³+3x²+3x–2
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=0 q=–3
So the new equation is:
t³+0t –3 [2]
As p=0, and q=–3
t= ∛(–q)=∛(3)
so x=t–a/3=∛(–q)–a/3=∛(3)–1
In summary, the roots for
Cubic: x³+3x²+3x–2 are:
x1=0.4422495703 (Remainder=0)
x2=–1.7211247852+1.2490247665i (Remainder=0)
x3=–1.7211247852–1.2490247665i (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+2x²+10x–20

In 1225, Leonardo da Pisa found the root 1.368808107 of this equation. For his time it is intriguing, especially because this a truncated value. Rounded to 9 places, it is  1.368808108.
Cubic in normal form: x³+2x²+10x–20
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=8.666666666666666 q=–26.074074074074076
So the new equation is:
t³+8.666666666666666t –26.074074074074076 [2]

The discriminant of the cubic, D=(p/3)³+(q/2)²
Before rounding, D=194.0740740740741
D=194.0740740740741

D> 0→ 2 complex roots
D=0 → real roots, with two the same
D<0 → Three real roots

Because D>0, two of the roots are complex

Our reduced equation is:
t³+8.666666666666666t –26.074074074074076 [2, repeated]
Or in algebra, t³+pt+q=0
There is a relationship:
(u–v)³+3uv(u–v)–(u³–v³)=0, which indicates how we might solve the equation
If we set t=u–v, p=3uv, or v=p/3u and q=–(u³–v³), that is:
v=8.666666666666666/(3u) [4]
u³–(8.666666666666666/(3*u)–26.074074074074076=0 [5]
So u³– p³/(27u³)+q= 0
Hence, We need to find numbers, u and v, such that:
u³–v³=–q, and uv=p/3, giving t=v–u
u³–v³=26.074074074074076, [6] and
uv=8.666666666666666/3 [7]
Using v=8.666666666666666/3u, we can eliminate u
u³– p³/27u³ =–q
The resulting equation is:
27u⁶ + 27qu³ –p³=0
Normalised, this is
u⁶ + qu³ –p³/27=0
=u⁶ +26.074074074074076*u³ –8.666666666666666³/27=0
This is a quadratic equation in u³, so we know how to solve it
u³=–q/2 ±√(q²/4 +p³/27) (5)
The square root may be negative, making the u's complex numbers, although the resulting roots might not be complex.
D=679.857338820302/4+650.9629629629628/27=
194.0740740740741
After rounding, D=194.0740740740741
The square root above isn't negative, so two of the roots are complex
u= ∛(–q/2+√(D))=2.9988174659594593
v= ∛(q/2+√(D))=0.9633426914714203
x1=u1–v1–a/3=1.3688081078213723,0
x2=–0.5*(u1–v1)–a/3+(u1+v1)*√(3)/2i
x2=–1.6844040539+3.4313313502i
Now we know x2, we also know that x3 simply has the opposite sign imaginary part:
So x3=Re(x2)–Im(x2)=–1.684404053910686–3.4313313501976923i, but here is the formula:
x3=–0.5*(u1–v1)–a/3+(u1+v1)*√(3)/2i
x3=–1.6844040539–3.4313313502i
In summary, the roots for
Cubic: x³+2x²+10x–20 are:
x1=1.3688081078 (Remainder=0)
x2=–1.6844040539+3.4313313502i (Remainder=0)
x3=–1.6844040539–3.4313313502i (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

 x³+x²+10x–3

Cubic in normal form: x³+x²+10x–3
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=9.666666666666666 q=–6.2592592592592595
So the new equation is:
t³+9.666666666666666t –6.2592592592592595 [2]

The discriminant of the cubic, D=(p/3)³+(q/2)²
Before rounding, D=43.249999999999986
D=43.249999999999986

D> 0→ 2 complex roots
D=0 → real roots, with two the same
D<0 → Three real roots

Because D>0, two of the roots are complex

Our reduced equation is:
t³+9.666666666666666t –6.2592592592592595 [2, repeated]
Or in algebra, t³+pt+q=0
There is a relationship:
(u–v)³+3uv(u–v)–(u³–v³)=0, which indicates how we might solve the equation
If we set t=u–v, p=3uv, or v=p/3u and q=–(u³–v³), that is:
v=9.666666666666666/(3u) [4]
u³–(9.666666666666666/(3*u)–6.2592592592592595=0 [5]
So u³– p³/(27u³)+q= 0
Hence, We need to find numbers, u and v, such that:
u³–v³=–q, and uv=p/3, giving t=v–u
u³–v³=6.2592592592592595, [6] and
uv=9.666666666666666/3 [7]
Using v=9.666666666666666/3u, we can eliminate u
u³– p³/27u³ =–q
The resulting equation is:
27u⁶ + 27qu³ –p³=0
Normalised, this is
u⁶ + qu³ –p³/27=0
=u⁶ +6.2592592592592595*u³ –9.666666666666666³/27=0
This is a quadratic equation in u³, so we know how to solve it
u³=–q/2 ±√(q²/4 +p³/27) (5)
The square root may be negative, making the u's complex numbers, although the resulting roots might not be complex.
D=39.17832647462277/4+903.2962962962962/27=
43.24999999999999
After rounding, D=43.24999999999999
The square root above isn't negative, so two of the roots are complex
u= ∛(–q/2+√(D))=2.133118405301851
v= ∛(q/2+√(D))=1.510568852724448
x1=u1–v1–a/3=0.28921621924406943,0
x2=–0.5*(u1–v1)–a/3+(u1+v1)*√(3)/2i
x2=–0.6446081096+3.1555257289i
Now we know x2, we also know that x3 simply has the opposite sign imaginary part:
So x3=Re(x2)–Im(x2)=–0.6446081096220346–3.1555257288964396i, but here is the formula:
x3=–0.5*(u1–v1)–a/3+(u1+v1)*√(3)/2i
x3=–0.6446081096–3.1555257289i
In summary, the roots for
Cubic: x³+x²+10x–3 are:
x1=0.2892162192 (Remainder=0)
x2=–0.6446081096+3.1555257289i (Remainder=0)
x3=–0.6446081096–3.1555257289i (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+4x²+x–6

Cubic in normal form: x³+4x²+x–6
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=–4.333333333333333 q=–2.5925925925925926
So the new equation is:
t³–4.333333333333333t –2.5925925925925926 [2]

The discriminant of the cubic, D=(p/3)³+(q/2)²
Before rounding, D=–1.3333333333333335
D=–1.333333333333333

D> 0→ 2 complex roots
D=0 → real roots, with two the same
D<0 → Three real roots

Because D<0, all the roots are real and distinct

Our reduced equation is:
t³–4.333333333333333t –2.5925925925925926 [2, repeated]
Or in algebra, t³+pt+q=0
There is a relationship:
(u–v)³+3uv(u–v)–(u³–v³)=0, which indicates how we might solve the equation
If we set t=u–v, p=3uv, or v=p/3u and q=–(u³–v³), that is:
v=–4.333333333333333/(3u) [4]
u³–(–4.333333333333333/(3*u)–2.5925925925925926=0 [5]
So u³– p³/(27u³)+q= 0
Hence, We need to find numbers, u and v, such that:
u³–v³=–q, and uv=p/3, giving t=v–u
u³–v³=2.5925925925925926, [6] and
uv=–4.333333333333333/3 [7]
Using v=–4.333333333333333/3u, we can eliminate u
u³– p³/27u³ =–q
The resulting equation is:
27u⁶ + 27qu³ –p³=0
Normalised, this is
u⁶ + qu³ –p³/27=0
=u⁶ +2.5925925925925926*u³ +4.333333333333333³/27=0
This is a quadratic equation in u³, so we know how to solve it
u³=–q/2 ±√(q²/4 +p³/27) (5)
The square root may be negative, making the u's complex numbers, although the resulting roots might not be complex.
D=6.72153635116598/4–81.37037037037035/27=
–1.3333333333333326
After rounding, D=–1.333333333333332
The square root is a complex number, because D<0
u=∛(1.2962962962962963 ±√(1.333333333333332) i)
We need to find the cube roots of u³= 1.2962962963+1.1547005384i

Convert to trigonometric form
D=(q/2)²+(p/3)³
So u³=–q/2+√(–D) i
As u³–v³=–q,
v³=q/2+√(–D) i
r is therefore the same for both u and v [A]
r²=(–q/2)²–D
=(–q/2)²–(q/2)²+(p/3)³)
=–(p/3)³)
r=√(–(p/3)³ )
Use cos to find φ, as this makes finding the angles easier (with a calculator or computer)
r=√( (4.333333333333333/3)³) (as it is a phasor, it can have positive values only.)
r=1.7360061696678466 and r^1/3=1.2018504251546631
cos (φ)=–q/2r
cos (φ)=2.5925925925925926/(2*1.7360061696678466)
For v, cos(φ_v)=–2.5925925925925926/(2*1.7360061696678466)
So, cos(φ_v)= – cos(φ) [B]
Similarly, sin(φ_v)= sin(φ) [C]
acos(φ)= –q/(2*r)
φ=0.7276916222864558
x1=u–v–a/3
u=r^1/3*(cos(φ/3) + i sin(φ/3)
v=r^1/3*(–cos(φ/3) + i sin(φ/3) (from equations B and C above)
So, x1=2*r*cos(φ)–a/3 (because the sines cancel and the cosines are doubled
x1=2*1.2018504251546631*cos(0.7276916222864558)–4/3
We find the other x's by adding 2π to φ, and dividing the result by 3
φ₂=φ+2*π
x2=2*1.2018504251546631*cos(2.3369589764886807)–4/3
φ₂=φ+4*π
x3=2*1.2018504251546631*cos(4.431354078881876)–4/3

In summary, the roots for
Cubic: x³+4x²+x–6 are:
x1=1 (Remainder=0)
x2=–3 (Remainder=0)
x3=–2 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

1000x³–1254x²–496x+191

Cubic in normal form: x³–1.254x²–0.496x+0.191
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=–1.020172 q=–0.16239726400000004
So the new equation is:
t³–1.020172t –0.16239726400000004 [2]

The discriminant of the cubic, D=(p/3)³+(q/2)²
Before rounding, D=–0.03273066871437038
D=–0.03273066871437

D> 0→ 2 complex roots
D=0 → real roots, with two the same
D<0 → Three real roots

Because D<0, all the roots are real and distinct

Our reduced equation is:
t³–1.020172t –0.16239726400000004 [2, repeated]
Or in algebra, t³+pt+q=0
There is a relationship:
(u–v)³+3uv(u–v)–(u³–v³)=0, which indicates how we might solve the equation
If we set t=u–v, p=3uv, or v=p/3u and q=–(u³–v³), that is:
v=–1.020172/(3u) [4]
u³–(–1.020172/(3*u)–0.16239726400000004=0 [5]
So u³– p³/(27u³)+q= 0
Hence, We need to find numbers, u and v, such that:
u³–v³=–q, and uv=p/3, giving t=v–u
u³–v³=0.16239726400000004, [6] and
uv=–1.020172/3 [7]
Using v=–1.020172/3u, we can eliminate u
u³– p³/27u³ =–q
The resulting equation is:
27u⁶ + 27qu³ –p³=0
Normalised, this is
u⁶ + qu³ –p³/27=0
=u⁶ +0.16239726400000004*u³ +1.020172³/27=0
This is a quadratic equation in u³, so we know how to solve it
u³=–q/2 ±√(q²/4 +p³/27) (5)
The square root may be negative, making the u's complex numbers, although the resulting roots might not be complex.
D=0.02637287135468571/4–1.0617449369321288/27=
–0.03273066871437038
After rounding, D=–0.03273066871437
The square root is a complex number, because D<0
u=∛(0.08119863200000002 ±√(0.03273066871437) i)
We need to find the cube roots of u³= 0.081198632+0.1809161925i

Convert to trigonometric form
D=(q/2)²+(p/3)³
So u³=–q/2+√(–D) i
As u³–v³=–q,
v³=q/2+√(–D) i
r is therefore the same for both u and v [A]
r²=(–q/2)²–D
=(–q/2)²–(q/2)²+(p/3)³)
=–(p/3)³)
r=√(–(p/3)³ )
Use cos to find φ, as this makes finding the angles easier (with a calculator or computer)
r=√( (1.020172/3)³) (as it is a phasor, it can have positive values only.)
r=0.1983025127249824 and r^1/3=0.5831443503398909
cos (φ)=–q/2r
cos (φ)=0.16239726400000004/(2*0.1983025127249824)
For v, cos(φ_v)=–0.16239726400000004/(2*0.1983025127249824)
So, cos(φ_v)= – cos(φ) [B]
Similarly, sin(φ_v)= sin(φ) [C]
acos(φ)= –q/(2*r)
φ=1.148924921167228
x1=u–v–a/3
u=r^1/3*(cos(φ/3) + i sin(φ/3)
v=r^1/3*(–cos(φ/3) + i sin(φ/3) (from equations B and C above)
So, x1=2*r*cos(φ)–a/3 (because the sines cancel and the cosines are doubled
x1=2*0.5831443503398909*cos(1.148924921167228)+1.254/3
We find the other x's by adding 2π to φ, and dividing the result by 3
φ₂=φ+2*π
x2=2*0.5831443503398909*cos(2.4773700761156046)––1.254/3
φ₂=φ+4*π
x3=2*0.5831443503398909*cos(4.571765178508801)––1.254/3

In summary, the roots for
Cubic: 1000x³–1254x²–496x+191 are:
x1=1.4997993055 (Remainder=0)
x2=–0.5003313644 (Remainder=0)
x3=0.254532059 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places




Ken Ward's Mathematics Pages

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Faster Arithmetic - by Ken Ward