# Find the cubic equation given the roots

• chwala
In summary, the conversation discusses different methods for finding the cubic equation ##x^3-3x^2+4=0##, with one method being to let ##u=2x## and simplifying the equation to ##u^3-6u^2+32=0##. The conversation also mentions replacing x with x/a and the potential confusion with using x as a variable. Finally, the conversation mentions finding the cubic equation given roots ##∝-2, β-2##, and ##γ-2## by setting ##u=x-2## and solving for the new equation.

#### chwala

Gold Member
Homework Statement
The cubic equation ##x^3-3x^2+4=0## has roots ##∝, β## and ##ϒ##. Find the cubic equation with roots ##2∝, 2β## and ##2ϒ##.
Relevant Equations
cubic equations with related roots
##\sum ∝=3##
##\sum ∝β=0##
##∝βγ=-4##
##\sum2 ∝=6##
##\sum 2∝.2β##=4##\sum ∝β=0##
##2∝.2β.2γ=-32##
we then end up with
##x^3-6x^2+0x+32=0##
##x^3-6x^2+32=0##

i am looking for alternative methods ...

the next part of the question is to find the cubic equation (for the same original problem) given the roots, ##∝-2, β-2## and ##γ-2##, i will attempt this later...once i look at other alternative methods...

You could also just let ##u=2x##, and simplify.

jbriggs444 and Delta2
What would happen to the roots if you replaced every x with x/a?

etotheipi said:
You could also just let ##u=2x##, and simplify.
not getting it...

chwala said:
not getting it...

##u=2x→ \frac {u}{2}=x##
##→\frac {u^2}{4}=x^2##
##→\frac {u^3}{8}=x^3##
substituting in the original equation gives;
##\frac {u^3}{8}-3\frac {u^2}{4}+4=0##
##u^3-6u^2+32=0## i still do not seem to get it...

chwala said:
##u^3-6u^2+32=0## i still do not seem to get it...

That's the equation you want, no?

etotheipi said:
That's the equation you want, no?

i thought the equation ought to be expressed in terms of ##x##...a bit confusing here

For a bit more context, you start with$$x^3 -3x^2 + 4 = 0$$suppose this is solved by ##x \in \{\xi_i\}##. Now, transform the equation with ##u = 2x##,$$\left(\frac{u}{2}\right)^3 -3\left(\frac{u}{2}\right)^2 + 4 = 0$$hopefully this makes it easier to see that ##u \in \{2\xi_i\}## will solve this new equation. You can rearrange this equation to get$$u^3 - 6u^2 + 32 = 0$$This is not the same equation you started with, it is a different equation with different solutions [which are double that of the previous equation]. But, you can just replace ##u## with ##x##, if you want to write the polynomial in terms of that symbol instead. Part of your confusion might be, that '##x##' is often overloaded.

Delta2 and chwala
etotheipi said:
For a bit more context, you start with$$x^3 -3x^2 + 4 = 0$$suppose this is solved by ##x \in \{\xi_i\}##. Now, transform the equation with ##u = 2x##,$$\left(\frac{u}{2}\right)^3 -3\left(\frac{u}{2}\right)^2 + 4 = 0$$hopefully this makes it easier to see that ##u \in \{2\xi_i\}## will solve this new equation. You can rearrange this equation to get$$u^3 - 6u^2 + 32 = 0$$This is not the same equation you started with, it is a different equation with different solutions [which are double that of the previous equation]. But, you can just replace ##u## with ##x##, if you want to write the polynomial in terms of that symbol instead. Part of your confusion might be, that '##x##' is often overloaded.

i see...

chwala said:
the next part of the question is to find the cubic equation (for the same original problem) given the roots, ##∝-2, β-2## and ##γ-2##, i will attempt this later...once i look at other alternative methods...
For this you set ##u=x-2## and proceed in a similar way as before. it will involve a bit more algebraic processing to find the final equation ##f(u)=0##.

etotheipi and chwala