Solving a cubic equation

[chwala]

Homework Statement

find $x$ if $2x(x^2y+4)=64$

Relevant Equations

cubic equations

$x^3y+4x-32=0$

is there a particular method for solving this? i know that $x=2$ and $y=3$

 

[WWGD]

I think you did not simplify correctly. I also assume you want integer solutions. Your 2 could have canceled out. After that, maybe you can solve for $x$ as a cubic and see what conditions on the solution produce integers.If I may, yours seems like unusual topics. Are you self -studying or taking a class?

 

[chwala]

I think you did not simplify correctly. I also assume you want integer solutions. Your 2 could have canceled out. After that, maybe you can solve for $x$ as a cubic and see what conditions on the solution produce integers.If I may, yours seems like unusual topics. Are you self -studying or taking a class?

i am coming up with my own problems actually. I create them...self- study mate

 

[chwala]

I think you did not simplify correctly. I also assume you want integer solutions. Your 2 could have canceled out. After that, maybe you can solve for $x$ as a cubic and see what conditions on the solution produce integers.If I may, yours seems like unusual topics. Are you self -studying or taking a class?

what do you mean by saying i did not simplify correctly? I think my equation is mathematically correct!

 

[WWGD]

what do you mean by saying i did not simplify correctly? I think my equation is mathematically correct!

Ah, I thought you had done something else. Are you familiar with the inverse or implicit function theorem? Edit: Or solve the cubic and see the conditions needed to find integer solutions.

 

[chwala]

Ah, I thought you had done something else. Are you familiar with the inverse or implicit function theorem? Edit: Or solve the cubic and see the conditions needed to find integer solutions.

i know implicit differentiation...guide me on the steps...

 

[WWGD]

I think my suggestion on differentiation may not have been the best. I think solving the cubic and finding conditions for integer solutions may be the best approach here. I need to be out for a while, will be back.

 

[chwala]

$3x^{2}+x^3y^{'}+4=0$
$x^{3}y^{'}=-4-3x^2y$

 

[chwala]

I think my suggestion on differentiation may not have been the best. I think solving the cubic and finding conditions for integer solutions may be the best approach here. I need to be out for a while, will be back.

ok i will also be awaiting more advice from other experts here...

 

[member 587159]

This equation has infinitely many solutions.

 

[archaic]

Rewrite it like this : $y(x^3+\frac{4}{y}x-\frac{32}{y})=0$

Derivation : https://en.wikipedia.org/wiki/Cubic_equation#Derivation_of_the_roots
For $y = 0$ you just need to solve $4x-32=0$.

 

[chwala]

Rewrite it like this : $y(x^3+\frac{4}{y}x-\frac{32}{y})=0$
View attachment 253569
Derivation : https://en.wikipedia.org/wiki/Cubic_equation#Derivation_of_the_roots
For $y = 0$ you just need to solve $4x-32=0$.

i will look at this, what's the name of the theorem? you guys are awesome

 

[chwala]

i will look at this, what's the name of the theorem? you guys are awesome

Bingo...

 

[SammyS]

Homework Statement:: find $x$ if $2x(x^2y+4)=64$
Homework Equations:: cubic equations

$x^3y+4x-32=0$

is there a particular method for solving this? i know that $x=2$ and $y=3$

Writing the equation in the form, $x(x^2y+4)=32$, shows that $x$ cannot be zero.

Solving the equation for $y$, shows that it's not so much like a cubic equation as it is a rational function, especially if you're looking for integer solutions.

$y=\dfrac{4(8-x)}{x^3}$

Just plug in a value for $x$, get a value for $y$.

 

[epenguin]

Yes, Your equation is just a relation between two variables, x and y.
Soution? Well, you could call the equation x = ... in #14 a solution.
You easily see from it that the relation between x and y it specifies is continuous except at x = 0, and that it covers an infinite range.
You have a simple relation y = a function of x in #14. You can work out a relation of form x = a function of y as told - but why bother? It is still saying the same thing in a more complicated way. in other cases you may not be able to work it out (cannot solve the equation algebraically) so you always go for the one that is simpler - in fact you are lucky that there is one, in many cases you can't get either an x =... or a y =... form, though you still have a relation which corresponds to a curve in two dimensions.

If you want whole number solutions, at least with small whole numbers, then I think the best way is simply to use a graphing calculator or app and see whether the graphed function goes through any small numbers. That immediately gave me your solution and also x = -2, y = -5.

I suspect you could prove without too much difficulty that there are no other whole number (er, integer) solutions but I am not personally tempted to make the effort.

 

[chwala]

Rewrite it like this : $y(x^3+\frac{4}{y}x-\frac{32}{y})=0$
View attachment 253569
Derivation : https://en.wikipedia.org/wiki/Cubic_equation#Derivation_of_the_roots
For $y = 0$ you just need to solve $4x-32=0$.

archaic thanks for this.

so is this the general way of solving the cubic equations that (do not have a quadratic factor) i.e $ax^3+bx+c=0$?... are they reffered to as depressed cubics? this is new to me...

 

[chwala]

I think my suggestion on differentiation may not have been the best. I think solving the cubic and finding conditions for integer solutions may be the best approach here. I need to be out for a while, will be back.

so this problem cannot be solved as per your approach- differentiation...

 

[WWGD]

so this problem cannot be solved as per your approach- differentiation...

No, I misunderstood your question. I thought you just wanted to express y as a function of x. It was like 2 a.m. and I could barely read my phone. I don't think it would take us too far.

 

[chwala]

No, I misunderstood your question. I thought you just wanted to express y as a function of x. It was like 2 a.m. and I could barely read my phone. I don't think it would take us too far.

ok no worries am liking it here, ...

 

[WWGD]

ok no worries am liking it here, ...

PF: Where nerds meet... ;).

 

[Buzz Bloom]

archaic thanks for this.

so is this the general way of solving the cubic equations that (do not have a quadratic factor) i.e.
ax^3+bx+c=0?
... are they referred to as depressed cubics? this is new to me...

Hi chwala:

This from Standard Mathematical Tables, Page 344 (1957 ed.).

y³ + py² + qy + r = 0​

Substituting y = x - p/3 yields

x³ + ax + b = 0, where​

a = (1/3)(3q - p²) and​

b = (1/27)(2p³ - 9pq + 27r)​

Regards,
Buzz