Solution to a sixth grade (degree 6) equation

[Ennio]

Homework Statement

can you plase support to solve this equation (sixth grade equation)?

Homework Equations

a*x^6 + b*x^4 - c = 0 with a,b,c>0

The Attempt at a Solution

t = x^2 --> t^3 + b/a t^2 - c/a = 0
variable change t = u - b/3a --> u^3 + p*u + q = 0 p,q>0
Cardano´s solution for a third grade equation?

 

[Mark44]

Homework Statement

can you plase support to solve this equation (sixth grade equation)?

Homework Equations

a*x^6 + b*x^4 - c = 0 with a,b,c>0

The Attempt at a Solution

t = x^2 --> t^3 + b/a t^2 - c/a = 0
variable change t = u - b/3a --> u^3 + p*u + q = 0 p,q>0
Cardano´s solution for a third grade equation?

Your question confused me at first, as I thought you were talking about the sixth grade in school. In English, your first equations would be classified as sixth degree and third degree, respectively.

I don't know of any way to solve the latter equation other than by using Cardano's method. In this case it is slightly simpler than for a general third degree equation, because there is no first-degree term. However, solving third degree equations using Cardano is not something that is taught in any U.S. grade school that I am aware of, and if taught at all at university level, very few of them would include it, to the best of my knowledge.

 

[RPinPA]

Your question confused me at first, as I thought you were talking about the sixth grade in school.

Me as well. But "degree" is "grado" in Italian and Spanish.

Cardano´s solution for a third [degree] equation?

Yes, Cardano's method would be appropriate. You appear to have transformed it into the form of equation (15) here, so you could proceed with the remainder of that calculation.
http://mathworld.wolfram.com/CubicFormula.html

What is your question for us?

 

[jim mcnamara]

I think your problem relates to names for grades in primary and secondary schools. The terms used in Britain/US do not match up. British terminology: 3rd form would be second year algebra (or 10-11 grade in US), 6th form in Britain would be doing what is college level math (freshman-sophmore) in the US, i.e., analysis, analytic geometry.

 

[Ennio]

Me as well. But "degree" is "grado" in Italian and Spanish.
Yes, Cardano's method would be appropriate. You appear to have transformed it into the form of equation (15) here, so you could proceed with the remainder of that calculation.
http://mathworld.wolfram.com/CubicFormula.html

What is your question for us?

Thnak you very much for the link. Yes Grado!

 

[Ennio]

I think your problem relates to names for grades in primary and secondary schools. The terms used in Britain/US do not match up. British terminology: 3rd form would be second year algebra (or 10-11 grade in US), 6th form in Britain would be doing what is college level math (freshman-sophmore) in the US, i.e., analysis, analytic geometry.

thanks for the clarification

 

[SammyS]

Homework Statement

can you plase support to solve this equation (sixth grade equation)?

Homework Equations

a*x^6 + b*x^4 - c = 0 with a,b,c>0

The Attempt at a Solution

t = x^2 --> t^3 + b/a t^2 - c/a = 0
variable change t = u - b/3a --> u^3 + p*u + q = 0 p,q>0
Cardano´s solution for a third grade equation?

Another way to get a deflated cubic (one having no quadratic term) as well as giving the solutions to $\ at^3 + b t^2 - c = 0\$ is as follows.

Variable change: $\displaystyle \ t = \frac{1}{w}\,.\$

Then we have

$\displaystyle \ a \left(\frac{1}{w^3}\right)+b \left(\frac{1}{w^2}\right) -c =0\,.\$​

Multiply by $\ w^3\$ and divide by $\ -c\$ and rearrange terms.

$\displaystyle w^3 -\frac{b}{c}w - \frac{a}{c} = 0$​

Also, you never indicated what sort of solutions you were interested in.

 

[SammyS]

t^3 + b/a t^2 − c/a = 0
variable change t = u − b/3a --> u^3 + p*u + q = 0 p,q>0
Cardano´s solution for a third grade equation?

I just noticed the restrictions that you have on p and q. Those are incorrect. In fact, it's definitely true that p < 0.
The sign of q depends upon the relative sizes of a, b, and c.

Spoiler: p & q

$\displaystyle p=-\, \frac { b^{2}}{3a^{2}} \$ and $\ \displaystyle q= \frac {2b^{3}}{27a^{3}} -\frac{c}{a}$