Roots of cubic equation

  • Thread starter seboastien
  • Start date
  • #1
53
0

Homework Statement



θ^3 - pθ^2 +qθ - r = 0 such that p and r do not equal zero

If the roots can be written in the form ak^-1, a, and ak for some constants a and k, show that one root is q/p and that q^3 - rp^3 = 0. Also, show that if r=q^3/p^3, show that q/p is a root and that the product of the other roots is (q/p)^2

Homework Equations





The Attempt at a Solution



Mind boggling, can anyone give me so much as a hint?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,325
1,008
...
If the roots can be written in the form ak^-1, a, and ak for some constants a and k,
...
What do you mean by ak^-1 ?

Do you mean a/k , OR do you mean 1/(ak) ?
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,325
1,008
If the roots are x , y, and z , then

(θ -x)(θ - y)(θ - z) = θ3 - pθ2 +qθ - r
 
  • #4
52
0
What do you mean by ak^-1 ?

Do you mean a/k , OR do you mean 1/(ak) ?
I am hazarding a guess that he means a/k, so that the roots form a geometric progression (which was usually the basis for a lot of 'roots of polynomial' questions I used to get)
 
  • #5
52
0
With a cubic expression in the form:
\begin{align}
ax^3 + bx^2 + cx + d \\
\end{align}
If we call the roots [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] then:
\begin{align}
a(x-\alpha)(x-\beta)(x-\gamma) = ax^3 + bx^2 + cx + d \\
\end{align}

Where:
\begin{align}
\alpha + \beta + \gamma = \frac{-b}{a} \\

\alpha\beta + \beta\gamma + \gamma\alpha= \frac{c}{a} \\

\alpha\beta\gamma= \frac{-d}{a} \\
\end{align}

In this specific case, a = 1. That lot should help you!
Also please notice that the 'a' I use to represent the coefficient of x cubed is not the same as the 'a' you have been given in the question.


Extra Hint (Only use if you are still stuck):




\begin{align}
\alpha\beta\gamma = \frac{a}{k} \times a \times ak = a^3
\end{align}
 
Last edited:

Related Threads on Roots of cubic equation

Replies
7
Views
3K
  • Last Post
Replies
15
Views
2K
Replies
10
Views
14K
Replies
6
Views
2K
Replies
4
Views
4K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
8K
Replies
2
Views
2K
Replies
14
Views
24K
Top