Proving bc=d for Cubic Function with r=-s

In summary, The conversation discusses a proof involving a cubic function with two unequal solutions and the relationship between its coefficients. The use of the complex conjugate root theorem and the factor theorem are suggested as methods to prove that bc=d. It is also mentioned that posting such problems in the Homework Help section can result in a faster response time.
  • #1
Aki
83
0
Hey I need help with this proof:

Consider the cubic function: x^3 + bx^2 + cx + d = 0 . If the two solutions of the cubic function are not equal, ie r != s, but r=-s, then prove that bc=d.

Thanks
 
Mathematics news on Phys.org
  • #2
I'll give you help on this one, but please, next time post these problems in the Homework Help section. Its not just for our sake, you'll get a faster reply there too.

-------------
If all coefficients are real, by the complex conjugate root theorem you know that since two of the roots are real, s and -s, then the third root must be real as well. So now use the factor theorem and write the polynomial as a product of its factors,

[tex]x^3+bx^2+cx+d = (x-k)(x^2-s^2)[/tex].

Equate coefficients and see what you get.
 
  • #3
Gib Z said:
I'll give you help on this one, but please, next time post these problems in the Homework Help section. Its not just for our sake, you'll get a faster reply there too.

-------------
If all coefficients are real, by the complex conjugate root theorem you know that since two of the roots are real, s and -s, then the third root must be real as well. So now use the factor theorem and write the polynomial as a product of its factors,

[tex]x^3+bx^2+cx+d = (x-k)(x^2-s^2)[/tex].

Equate coefficients and see what you get.


Very helpful. Thanks a lot Gib Z
 

FAQ: Proving bc=d for Cubic Function with r=-s

What does it mean to prove bc=d for a cubic function with r=-s?

Proving bc=d for a cubic function with r=-s means showing that the product of the coefficients b and c of the quadratic term in the function is equal to the constant term d, when the cubic function is written in the form ax^3+bx^2+cx+d and the coefficient of the linear term is equal to the negative of the coefficient of the quadratic term (r=-s).

Why is it important to prove bc=d for a cubic function with r=-s?

Proving bc=d for a cubic function with r=-s is important because it is a crucial step in solving cubic equations. By proving this relationship, we can use the quadratic formula to find the roots of the cubic function, making it easier to solve and analyze.

How is proving bc=d for a cubic function with r=-s different from other methods of solving cubic equations?

Proving bc=d for a cubic function with r=-s is different from other methods of solving cubic equations because it relies on finding the relationship between the coefficients b, c, and d, rather than directly finding the roots of the equation. This method is often easier and more efficient than other methods.

What are the steps involved in proving bc=d for a cubic function with r=-s?

The steps involved in proving bc=d for a cubic function with r=-s are: 1) Rewrite the cubic function in the form ax^3+bx^2+cx+d, with r=-s; 2) Multiply the coefficients b and c of the quadratic term, and equate it to the constant term d; 3) Solve for the unknown variable using algebraic techniques; 4) Substitute the value found into the cubic function and simplify to show that bc=d.

Can proving bc=d for a cubic function with r=-s be applied to all cubic functions?

Yes, proving bc=d for a cubic function with r=-s can be applied to all cubic functions, as long as they can be written in the form ax^3+bx^2+cx+d and the coefficient of the linear term is equal to the negative of the coefficient of the quadratic term (r=-s). This method works for both real and complex roots.

Similar threads

Replies
16
Views
4K
Replies
4
Views
2K
Replies
4
Views
11K
Replies
4
Views
2K
Replies
1
Views
835
Back
Top