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Cardano's Method to Solve Cubic Equations

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

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

electronics concepts will be posted in a week or two!

Yes Friends!

The author of this blog will be starting his preparation for gate in the may and June and will be revising all subjects what he has studied.

This is the first time ever that he is getting a long break in his engineering....

As being a regular student he was not able to go deep because his exams do not need depth and concepts...

The subjects he will be covering are as follows:-

1)electronic devices
2)network theory
3)Mathematics
4)probability and random processes
5)signals and systems
6)analog circuits
7)electromagnetics
8)control systems
9)digital electronics and logic design
10)analog communication

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