Developing a formula from cube equation

so in summary, the formulas for r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, and r_{1}r_{2}r_{3} are:r_{1}+r_{2}+r_{3}=-ar_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}=br_{1}r_{2}r_{3}=-c
  • #1
elimenohpee
67
0

Homework Statement


Given that the equation [tex] x^{3} + ax^{2} + bx + c = 0 [/tex] has the roots [tex]r_{1}, r_{2}, r_{3}[/tex], develop formulae for [tex]r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3} [/tex]


Homework Equations





The Attempt at a Solution



Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
[tex](x-r_{1})(x-r_{2})(x-r_{3})=0[/tex]
 
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  • #2
elimenohpee said:

Homework Statement


Given that the equation [tex] x^{3} + ax^{2} + bx + c = 0 [/tex] has the roots [tex]r_{1}, r_{2}, r_{3}[/tex], develop formulae for [tex]r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3} [/tex]


Homework Equations





The Attempt at a Solution



Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
[tex](x-r_{1})(x-r_{2})(x-r_{3})=0[/tex]

Multiply those out and see what happens.
 
  • #3
Is that really all I need to do? I mean when I multiply it out, I get:

[tex] x^{3}-cx^{2}-bx^{2}-ax^{2}+bcx+acx+abx-abc=0[/tex]

The question says to develop formulae, all three formulas are within this equation...is there anything else I can really say here? lol
 
  • #4
elimenohpee said:

Homework Statement


Given that the equation [tex] x^{3} + ax^{2} + bx + c = 0 [/tex] has the roots [tex]r_{1}, r_{2}, r_{3}[/tex], develop formulae for [tex]r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3} [/tex]


Homework Equations





The Attempt at a Solution



Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
[tex](x-r_{1})(x-r_{2})(x-r_{3})=0[/tex]

LCKurtz said:
Multiply those out and see what happens.

elimenohpee said:
Is that really all I need to do? I mean when I multiply it out, I get:

[tex] x^{3}-cx^{2}-bx^{2}-ax^{2}+bcx+acx+abx-abc=0[/tex]

The question says to develop formulae, all three formulas are within this equation...is there anything else I can really say here? lol

? When you multiply out

[tex](x-r_{1})(x-r_{2})(x-r_{3})=0[/tex]

you won't have any a, b, or c. The question is how does what you get relate to the original cubic you started with:

[tex]
x^{3} + ax^{2} + bx + c = 0
[/tex]
 
  • #5
wow, sorry I see now. Forgive what I just wrote, it is late and I wasn't thinking correctly.

When I multiply out, I get:

[tex] x^{3} -x^{2}(r_{1}+r_{2}+r_{3}) + x(r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}) -r_{1}r_{2}r_{3}=0 [/tex]

then you can equate coefficients with the original cubic expression to get:

[tex]x^{3}[/tex] term:[tex] 1-1=0 [/tex]
[tex]x^{2}[/tex] term: [tex] -(r_{1}+r_{2}+r_{3})=a [/tex]
[tex]x[/tex] term:[tex] r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}=b [/tex]
initial term:[tex] r_{1}r_{2}r_{3} = -c [/tex]
 

1. How do you develop a formula from a cube equation?

To develop a formula from a cube equation, you need to begin by understanding the basic structure of the equation. A cube equation is in the form of ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants. From this form, you can use the quadratic formula to solve for x, and then substitute that value into the equation to solve for the other constants. This will give you a general formula for all solutions to the cube equation.

2. What is the purpose of developing a formula from a cube equation?

The purpose of developing a formula from a cube equation is to find a general solution that will work for any given cube equation. This formula can then be used to solve for any specific values of a, b, c, and d, making it a powerful tool for solving problems in mathematics and science.

3. Are there any limitations to the formula developed from a cube equation?

Yes, there are limitations to the formula developed from a cube equation. It will only work for cube equations that are in the form of ax^3 + bx^2 + cx + d = 0. Additionally, it will only give you real solutions, so if the equation has complex solutions, they will not be included in the formula.

4. Can the formula developed from a cube equation be used for higher degree equations?

No, the formula developed from a cube equation can only be used for cube equations. For equations with higher degrees, such as quartic or quintic equations, different methods and formulas are needed to find solutions.

5. How can the formula developed from a cube equation be used in real-world applications?

The formula developed from a cube equation can be used in various real-world applications, such as predicting trends in data or modeling physical systems. It can also be used in the fields of engineering, physics, and economics to solve problems and make calculations more efficient.

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