12000=2200a+4840a^2+7986a^3
solve this equation for value of a :afrog:
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12000=2200a+4840a^2+7986a^3
solve this equation for value of a :afrog:
try this
http://www.1728.com/cubic.htm
http://www.hvks.com/Numerical/websolver.htmQuote:
Originally Posted by dino_lawki
Quote:
For the real Polynomial:
+7986x^3+4840x^2+2200x-12000
Start Newton Iteration for Polynomial=+7986x^3+4840x^2+2200x-12000
Stage 1=>Stop Condition: f(z)<1.13e-11
Start : z[1]=(0.6+i0) dz=(-5.73e-1+i0.00e+0) f(z)=7.7e+3
Iteration: 1
Newton Step: z[1]=(1+i0) dz=(-4.91e-1+i0.00e+0) f(z)=5.4e+3
Function value decrease=>try multiple steps in that direction
Try Step: z[1]=(2+i0) dz=(-4.91e-1+i0.00e+0) f(z)=3.3e+4
: No improvement=>Discard last try step
Iteration: 2
Newton Step: z[1]=(0.9+i0) dz=(1.37e-1+i0.00e+0) f(z)=5.5e+2
Enter Stage 2=>New Stop Condition: f(z)<6.99e-11
Iteration: 3
Newton Step: z[1]=(0.9+i0) dz=(1.72e-2+i0.00e+0) f(z)=8.0e+0
Enter Stage 2=>New Stop Condition: f(z)<6.62e-11
Iteration: 4
Newton Step: z[1]=(0.9+i0) dz=(2.58e-4+i0.00e+0) f(z)=1.8e-3
Enter Stage 2=>New Stop Condition: f(z)<6.62e-11
Iteration: 5
Newton Step: z[1]=(0.9+i0) dz=(5.76e-8+i0.00e+0) f(z)=8.4e-11
Enter Stage 2=>New Stop Condition: f(z)<6.62e-11
Iteration: 6
Newton Step: z[1]=(0.9+i0) dz=(2.72e-15+i0.00e+0) f(z)=5.5e-12
Enter Stage 2=>New Stop Condition: f(z)<6.62e-11
Stop Criteria satisfied after 6 Iterations
Final Newton z[1]=(0.9+i0) dz=(2.72e-15+i0.00e+0) f(z)=5.5e-12
Deflate the real root z=0.9090909090909093
Solved Polynomial=+7986x^2+12100.000000000001x+13200.000000000003 directly
The Solutions are:
X1=(-0.7575757575757577+i1.0387355455153097)
X2=(-0.7575757575757577-i1.0387355455153097)
X3=0.9090909090909093
i took a stab at in using the synthetic division method without any luck. Although, I didn't really try hard still might be a real number answer.
http://mathworld.wolfram.com/CubicFormula.htmlQuote:
Originally Posted by superbovine
The exact real solution is a = 10/11 (found with a HP 15-C handheld calculator)
nice