Solving a Cubic Equation: Methods and Terminology

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In summary: 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 long to solve this with the method I suggested, but I don't see myself having the time to do it anytime soon. What I meant was that solving the cubic may give us insight into how to find integer solutions.
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chwala
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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##
 
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  • #2
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?
 
  • #3
WWGD said:
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
 
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  • #4
WWGD said:
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!
 
  • #5
chwala said:
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.
 
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  • #6
WWGD said:
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...
 
  • #7
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.
 
  • #8
##3x^{2}+x^3y^{'}+4=0##
##x^{3}y^{'}=-4-3x^2y##
 
  • #9
WWGD said:
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...
 
  • #10
This equation has infinitely many solutions.
 
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  • #11
Rewrite it like this : ##y(x^3+\frac{4}{y}x-\frac{32}{y})=0##
Capture.PNG

Derivation : https://en.wikipedia.org/wiki/Cubic_equation#Derivation_of_the_roots
For ##y = 0## you just need to solve ##4x-32=0##.
 
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  • #13
chwala said:
i will look at this, what's the name of the theorem? you guys are awesome:wink:

Bingo...
 
  • #14
chwala said:
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##.
 
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  • #15
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.
 
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  • #17
WWGD said:
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...
 
  • #18
chwala said:
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.
 
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  • #19
WWGD said:
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, ...
 
  • #20
chwala said:
ok no worries am liking it here, ...
PF: Where nerds meet... ;).
 
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  • #21
chwala said:
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
 
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Related to Solving a Cubic Equation: Methods and Terminology

1. What is a cubic equation?

A cubic equation is a polynomial equation of the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants and x is the variable. It is called a cubic equation because the highest degree of the variable is 3.

2. How do you solve a cubic equation?

To solve a cubic equation, you can use the cubic formula or the factor theorem. The cubic formula is a general formula that can be used to find the solutions of any cubic equation. The factor theorem involves factoring the equation into simpler expressions and solving for the variable.

3. Can all cubic equations be solved?

Yes, all cubic equations have three solutions, which may be real or complex. However, some solutions may not be easily expressed in terms of real numbers. In these cases, approximations or numerical methods may be used to find the solutions.

4. What is the significance of solving a cubic equation?

Solving a cubic equation is important in many fields, including mathematics, physics, and engineering. It allows us to find the roots or solutions of the equation, which can help us understand the behavior of the system described by the equation. Cubic equations are also used in many applications, such as in designing bridges and analyzing population growth.

5. Are there any tips for solving a cubic equation?

One tip for solving a cubic equation is to first check if it can be factored. If it can be factored, it will be easier to find the solutions. Another tip is to use the rational root theorem, which states that the possible rational roots of a polynomial equation are all the factors of the constant term divided by all the factors of the leading coefficient. This can help narrow down the possible solutions and make the solving process more efficient.

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