# The best way to solve x³ + bx = c

## Main Question or Discussion Point

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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u mean 606087.936...? or 606087936

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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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

now I look at my answer I got confused... cause the formula does not contain b so it is wrong???

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D H
Staff Emeritus
Your result, $x=\exp((\log c)/3)$ is a convoluted way of writing $x=\sqrt[3]c$.

What you did with logarithms is wrong.

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Mentallic
Homework Helper
ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c
$\ln(x+ib)$ should be $$\ln\sqrt{x^2+b^2}+i\cdot \arg(z)$$

Because for a complex number x+iy, we can convert it into the mod-arg form $$r e^{i\theta}$$ and then $$\ln\left(r\cdot e^{i\theta}\right)=\ln(r)+i\theta$$ where $$r=\sqrt{x^2+y^2}, \theta=\tan^{-1}\left(\frac{y}{x}\right)$$

HallsofIvy
Homework Helper
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, $(a- b)^3= a^3- 3a^2b+ 3ab^2- b^3$ and $3ab(a- b)= 3a^2b- 3ab^2$ so that $(a- b)^3+ 3ab(a- b)= a^3- b^3$. That is, if x=a- b, m= 3ab, $n= a^3- b^3$, $x^3+ mx=n$.

Now that will fit $x^3+ 7x= 606,087.936$ with 3ab= 7, $a^3- b^3= 606087.936. Of course, that means b=7/a so [itex]a^3- 7^3/a^3= 606087.936$. Multiplying through by $a^3$, gives $(a^3)^2- 606087.936a^3- 7^3= 0$, a quadratic function for $a^3$. We can solve for $a^3$ using the quadratic formula:
$$a^3= \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3)}}{2}$$

So computing $a^3$ involves:
1) Find $(606087.936)^2$- one operation
2) Find $4(7^3)$- three operations
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 $b^3$, we have $a^3- b^3= 606087.936$ so $b^3= a^3- 606087.936$, 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.

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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$\ln(x+ib)$ should be $$\ln\sqrt{x^2+b^2}+i\cdot \arg(z)$$
Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354

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Mentallic
Homework Helper
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.

...We can solve for $a^3$ using the quadratic formula:
$$a^3 = \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3 ?)}}{2}$$
a) Probably that is a typo, shouldn't it be 7³/ 27 ?

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

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? :
$\ x^5 + bx^3 = c ,... x^5 + bx = c,... x^5 + bx^3 + cx =d$, .....

*should I make a new thread for this question?
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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HallsofIvy
Homework Helper
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.

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
$\frac{\sqrt[3]{2b/3}}{\sqrt[3]{\sqrt{3}\sqrt{4b^3+27c^2-9c}}}$ ....
is that any better?

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