Re: Cubic equations again...

• To: mathgroup at smc.vnet.net
• Subject: [mg115495] Re: Cubic equations again...
• From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
• Date: Thu, 13 Jan 2011 03:23:19 -0500 (EST)

```Phil,

there is something wrong with your calculation. I guess, it is the use of the parameter K. I would better take a
low case letter k. In addition you made one wrong conclusion.

Let us do it. I first simplify your equation by renaming:

expr = Simplify[(a - k -
2 (1 + b) q - (Sqrt[k] Sqrt[g])/Sqrt[q] /. {q ->  x^2, g ->  G/k,
a ->  k + A, b ->  c/2 - 1}), Assumptions ->  {x>  0, k>  0, g>  0}]
(*Assumptions x>0, just to look for a most simple case first, while k,g>0, since you later gave them positive values *)

A - (Sqrt[G] + c x^3)/x

Here are your three solutions, now in a form that you can more easily look at:

Clear[A, G, c];
sol1 = Solve[expr == 0, x][[1, 1, 2]]
sol2 = Solve[expr == 0, x][[2, 1, 2]]
sol3 = Solve[expr == 0, x][[3, 1, 2]]

-(((2/3)^(1/3)
A)/(9 c^2 Sqrt[G] + Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(
1/3)) - (9 c^2 Sqrt[G] + Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(
1/3)/(2^(1/3) 3^(2/3) c)

((1 + I Sqrt[3]) A)/(
2^(2/3) 3^(
1/3) (9 c^2 Sqrt[G] + Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(
1/3)) + ((1 - I Sqrt[3]) (9 c^2 Sqrt[G] +
Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(1/3))/(2 2^(1/3) 3^(2/3) c)

((1 - I Sqrt[3]) A)/(
2^(2/3) 3^(
1/3) (9 c^2 Sqrt[G] + Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(
1/3)) + ((1 + I Sqrt[3]) (9 c^2 Sqrt[G] +
Sqrt[3] Sqrt[-4 A^3 c^3 + 27 c^4 G])^(1/3))/(2 2^(1/3) 3^(2/3) c)

The two of them seem to be complex, while one seems to be real, but this is not necessarily the case.
Let us take your values of the parameters and substitute ->  first into the new parameters and then
into the obtained solutions:

A = a - k /. {a ->  10, b ->  3, k ->  0.1, g ->  1}
G = g*k /. {a ->  10, b ->  3, k ->  0.1, g ->  1}
c = 2 (1 + b) /. {a ->  10, b ->  3, k ->  0.1, g ->  1}

9.9

0.1

8

x1 = sol1 // Chop
x2 = sol2 // Chop
x3 = sol3 // Chop

-1.12807

1.0961

0.0319686

You get, therefore, all three solutions real, rather than one real and two complex. Legal are however,
only the two last, since we used the assumption x>0, so that the first should be additionally checked.

One way is to evaluate the following:

Plot[expr // N // Chop, {x, -1.13, -1.12}]
Plot[expr // N // Chop, {x, 1.09, 1.1}]
Plot[expr // N // Chop, {x, 0.02, 0.035}]

Solutions you want are then

q1 = x1^2
q2 = x2^2
q3 = x3^2

1.27254

1.20144

0.00102199

Have fun, Alexei

Related questions have been asked here before but I do not see how
they apply to the following problem.

I have solved a cubic equation using Solve to get an analytic
solution.  In the case I am dealing with I am interested in a real
root.

I then plug in the actual values of coefficients into the real
solution generated and get an answer that is incorrect.

Here are the details:

In[14]:= Solve[
a - K - 2 (1 + b) q - (Sqrt[K] Sqrt[zhi])/Sqrt[q] == 0, q]

Out[14]....presents three solutions, the first I simplify and then
assign to a function solq:

solq[a_, b_, K_, zhi_]=(a (1 + b) - (1 + b) K + (3 Sqrt[3]
Sqrt[(1 + b)^7 K zhi (-2 (a - K)^3 + 27 (1 + b) K zhi)] + (1 +
b)^3 (-(a - K)^3 + 27 (1 + b) K zhi))^(
1/3))^2/(6 (1 +
b)^2 (3 Sqrt[3]
Sqrt[(1 + b)^7 K zhi (-2 (a - K)^3 + 27 (1 + b) K zhi)] + (1 +
b)^3 (-(a - K)^3 + 27 (1 + b) K zhi))^(1/3))

Evaluating: solq[10, 3, .1, 1]
I get 1.2725408113389869 + 0. I  ( a real number).

Now evaluating

In[35]:= a - K - 2 (1 + b) q - (Sqrt[K] Sqrt[zhi])/Sqrt[
q] /. {a ->  10, b ->  3, K ->  .1, zhi ->  1, q ->
1.2725408113389869` }

Out[35]= -0.5606529814237877

So the solution is wrong.

The actual solution is q=1.2014371972809619

which can be confirmed by graphing or numerical solution using
FindRoot

In[12]:= Rt[a_, b_, K_, zhi_] :=
FindRoot[a - K - 2 (1 + b) q - (Sqrt[K] Sqrt[zhi])/Sqrt[q], {q, 0.1}]

In[13]:= Rt[10, 3, .1, 1]

Out[13]= {q ->  1.2014371972809619}

Does Solve not find the root, or is the problem evaluation of the
analytical expression with the selected values?

Thank you in advance,

Phil
University of Rochester

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Alexei Boulbitch, Dr. habil.
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