The best way to solve x³ + bx = c

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In summary, the conversation discusses the quickest way to solve a Khayyàm equation when b is not a square number. It is mentioned that x³ + 7x = 606,087.936 and x³ + 5x = 132906... are examples of such equations. The conversation then delves into finding the solution using the best algorithm and how many operations are required. It is also questioned whether it is possible to find the original equation or if there are multiple equations that can have the same solution. The method used is referred to as "Cardano's cubic formula" and it is discussed how it can be applied to equations with negative values for b and if it gives one of the solutions. The conversation
  • #1
logics
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What is the quickest way to solve a Khayyàm equation when b is not a square number?

Consider: x³ + 7x = 606,087.936,
( x³ + 5x = 132906...)

using the best algorithm, how many operations are required to find the solution ( x = 84.6, 51... ) ?
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
 
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  • #2
u mean 606087.936...? or 606087936
 
  • #3
n_kelthuzad said:
u mean 606087.936...? or 606087936
Thanks, Victor, I corrected the typo, but, 846 or 84.6, I suppose the operations do not change.
 
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  • #4
x^3+bx=c
x(x^2+b)=c
x(x+ib)(x-ib)=c
ln (x(x+ib)(x-ib))=ln c
ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c
ln x + ln x + ln x = ln c
3ln x = ln c
ln x = (ln c)/3
x=e^(ln c)/3
 
  • #5
now I look at my answer I got confused... cause the formula does not contain b so it is wrong?
 
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  • #6
Your result, [itex]x=\exp((\log c)/3)[/itex] is a convoluted way of writing [itex]x=\sqrt[3]c[/itex].

What you did with logarithms is wrong.
 
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  • #7
n_kelthuzad said:
ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c

[itex]\ln(x+ib)[/itex] should be [tex]\ln\sqrt{x^2+b^2}+i\cdot \arg(z)[/tex]

Because for a complex number x+iy, we can convert it into the mod-arg form [tex]r e^{i\theta}[/tex] and then [tex]\ln\left(r\cdot e^{i\theta}\right)=\ln(r)+i\theta[/tex] where [tex]r=\sqrt{x^2+y^2}, \theta=\tan^{-1}\left(\frac{y}{x}\right)[/tex]
 
  • #8
logics said:
What is the quickest way to solve a Khayyàm equation when b is not a square number?

Consider: x³ + 7x = 606,087.936,
( x³ + 5x = 132906...)

using the best algorithm, how many operations are required to find the solution ( x = 84.6, 51... ) ?
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
For any numbers, a and b, [itex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex] and [itex]3ab(a- b)= 3a^2b- 3ab^2[/itex] so that [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]. That is, if x=a- b, m= 3ab, [itex]n= a^3- b^3[/itex], [itex]x^3+ mx=n[/itex].

Now that will fit [itex]x^3+ 7x= 606,087.936[/itex] with 3ab= 7, [itex]a^3- b^3= 606087.936.

Of course, that means b=7/a so [itex]a^3- 7^3/a^3= 606087.936[/itex]. Multiplying through by [itex]a^3[/itex], gives [itex](a^3)^2- 606087.936a^3- 7^3= 0[/itex], a quadratic function for [itex]a^3[/itex]. We can solve for [itex]a^3[/itex] using the quadratic formula:
[tex]a^3= \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3)}}{2}[/tex]

So computing [itex]a^3[/itex] involves:
1) Find [itex](606087.936)^2[/itex]- one operation
2) Find [itex]4(7^3)[/itex]- three operations
3) Add them- one operation
4) Take the square root- one operation
5) Add to or subtract from 606087.936- one operation
6) Divide by 2- one operation
7) Take the cube root

That is, 8 operations are required to find just a. We don't have to repeat all that to find [itex]b^3[/itex], we have [itex]a^3- b^3= 606087.936[/itex] so [itex]b^3= a^3- 606087.936[/itex], a ninth operation, and then we must take the cube root, a tenth operation, to find b itself. Finally, x=a- b after a total of eleven operations.
 
  • #9
HallsofIvy said:
That is, 8 operations are required . Finally, x=a- b after a total of eleven operations.
Thanks a lot, Sir.
What is the name of this method?, only 11, actually 9 [if we consider 7³ just one] operations is definitely an eccellent achievement, it should apply also to equations where b is negative, in the region where there is only one solution i.e.:
x³ -9x = 148400 (x = 53, c> 10,3...) or maybe even when x > 3.
Can we say this algorithm gives always the or one of the solutions of a depressed cubic?
 
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  • #10
Mentallic said:
[itex]\ln(x+ib)[/itex] should be [tex]\ln\sqrt{x^2+b^2}+i\cdot \arg(z)[/tex]
Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354
 
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  • #11
logics said:
Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354

No, it doesn't get us any closer to solving the problem at hand. I was just showing n_kelthuzad why his solution was invalid.
 
  • #12
HallsofIvy said:
...We can solve for [itex]a^3[/itex] using the quadratic formula:
[tex]a^3 = \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3 ?)}}{2}[/tex]
a) Probably that is a typo, shouldn't it be 7³/ 27 ?

[itex]a^3 = \left(\frac{x + \sqrt{x^2 + 4*b}}{2}\right)^3 = \frac{c+\sqrt{c^2 + 4 * b^3/3^3}}{2}[/itex]

b) What happens if b is negative: x³ - bx = c, how do we find the other two solutions?

c) * Is there a similar brilliant solution for quintic equations with only one solution? :
[itex]\ x^5 + bx^3 = c ,... x^5 + bx = c,... x^5 + bx^3 + cx =d[/itex], ...

*should I make a new thread for this question?
logics said:
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
Can we find an equivalent original equation x³ + bx² +cx... having only one solution : 84.6?, does it depend on the relation between b and c?.
 
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  • #13
logics said:
Thanks a lot, Sir.
What is the name of this method?
That is "Cardano's cubic formula"
http://en.wikipedia.org/wiki/Cubic_function

only 11, actually 9 [if we consider 7³ just one] operations is definitely an eccellent achievement, it should apply also to equations where b is negative, in the region where there is only one solution i.e.:
x³ -9x = 148400 (x = 53, c> 10,3...) or maybe even when x > 3.
Can we say this algorithm gives always the or one of the solutions of a depressed cubic?
I don't know what you could mean by "the" solution of a cubic.
 
  • #14
HallsofIvy said:
I don't know what you could mean by "the" solution of a cubic.
I consider the question from the point of view of logics:
If, [on a "function explorer"], you look at the curve x³+ bx = c you see a curve similar, almost identical, to x³= c.. Could one ever say that x³ = c has one real and two imaginary solutions? Am I missing something?
I suppose that if, historically, "casus irriducibilis" had not occurred, we would not be here to talk about imaginary solutions (or numbers).

I know the fundamental theorem says "nth power-n zeroes", but I suppose it is a general frame and it is understood "at most" with b> 0, and only odd powers : x³+x;.. x^5 + x³+ x, etc..., there is only one solution, am I wrong?.
That is why I ventured to say that x³ + bx= 0 is a cubic equation with only one solution. How should one identify these equations?
On the other hand, also a cubic with b<0 or with even power has 2/more roots in an infinitesimal region of the y-axis, elsewhere we find the solution.
I hope I made myself clear.
Is there an algorithm to solve a depressed quintic x^5+... when there is only one solution?

P.S: at wolfram, sometimes they show the algorithm, a formula starting with
[itex]\frac{\sqrt[3]{2b/3}}{\sqrt[3]{\sqrt{3}\sqrt{4b^3+27c^2-9c}}}[/itex] ...
is that any better?
 
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1. What is the most efficient method for solving x³ + bx = c?

The most efficient method for solving this equation is by using the cubic formula, which is derived from the quadratic formula. It involves finding the roots of a cubic polynomial equation, which can be complex and time-consuming.

2. Can x³ + bx = c be solved using algebraic methods?

Yes, x³ + bx = c can be solved using algebraic methods such as factoring, completing the square, or using the quadratic formula. However, these methods may not always yield exact solutions and may require approximations.

3. Is there a simpler way to solve x³ + bx = c?

Unfortunately, there is no simpler way to solve x³ + bx = c than using the cubic formula. This equation is a third-degree polynomial, and there is no general formula for solving higher-degree polynomials.

4. Can x³ + bx = c be solved using numerical methods?

Yes, x³ + bx = c can be solved using numerical methods such as Newton's method or the bisection method. These methods involve using a series of approximations to find the roots of the equation.

5. Are there any special cases for solving x³ + bx = c?

Yes, there are a few special cases for solving x³ + bx = c. For example, if b = 0, the equation simplifies to x³ = c, which can be solved using cube roots. Additionally, if c = 0, the equation becomes x³ + bx = 0, which can be factored and solved for the roots.

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