Find the cubic equation given the roots

[chwala]

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 ...

 

[chwala]

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...

 

[etotheipi]

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

 

[Vanadium 50]

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

 

[chwala]

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

not getting it...

 

[chwala]

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...

 

[etotheipi]

$u^3-6u^2+32=0$ i still do not seem to get it...

That's the equation you want, no?

 

[chwala]

That's the equation you want, no?

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

 

[etotheipi]

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.

 

[chwala]

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...

 

[Delta2]

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 $f(u)=0$.