Polynomials (2 Viewers)

[Lukybear]

If w is a compelx cube root of 1, form a cubic with root (a+b)^-1, (aw+b)^-1, (aw^2 + b)^-1

 

[pwoh]

a and b are just arbitrary constants right?

Btw I have no idea, I hope someone answers this..

 

[Trebla]

If w is a compelx cube root of 1, form a cubic with root (a+b)^-1, (aw+b)^-1, (aw^2 + b)^-1

If w is a complex root of a cubic polynomial of unity (x³ - 1 = 0) then so is 1 and w² i.e. the polynomial has roots 1, w and w². It might make it easier to picture by calling them α, β and γ which means you want an equation with roots (aα + b)^-1, (aβ + b)^-1 and (aγ + b)^-1 which is now a standard problem of forming polynomial equations.

Let y = (ax + b)^-1, make x the subject and sub it into the polynomial equation.

 

[Lukybear]

What about this: Find the conditions for the roots of the cubic equation ax^3 + bx^2 +cx + d = 0 to be in geometric sequence

 

[Lukybear]

And also prove that

if w is a root of P(z) = z^8-5/2z^4+1
then 1/w is also a root

Also find one of the roots of P(z)=0 in exact form

 

[omgmynameislong]

sub in w and then sub in 1/w. For the 1/w equation times the whole thing by w^8 then ull get the same equation.

 

[shaon0]

And also prove that

if w is a root of P(z) = z^8-5/2z^4+1
then 1/w is also a root

Also find one of the roots of P(z)=0 in exact form

P(z)=(z^4-5/4)^2-(3/4)^2
=(z^4-2)(z^4-1/2)

For roots to be in GS:
Let roots be; A/r,A,Ar.
And you'll get A=cbrt(-d/a) [Using sum of roots]
If this is not enough;
r+r^-1=b^3/(a^2.d)-1 [Using prod and sum of roots]
And; A=sqrt((adc/(b^3-a^2.d)))
And; A=-b/c [Using sum of roots twice and sum of roots]
This question seems pretty vague but the conditions above (if they're correct) are probs enough unless they're asking for a Geometric argument.

 

[nrlwinner]

What about this: Find the conditions for the roots of the cubic equation ax^3 + bx^2 +cx + d = 0 to be in geometric sequence

Let roots be

Using product of roots...





Subbing it back into the equation, as a is a root







Then cubing








Hopefully that's right.

 

[MOP777]

And also prove that

if w is a root of P(z) = z^8-5/2z^4+1
then 1/w is also a root

Also find one of the roots of P(z)=0 in exact form

Isn't it quicker for this equation to prove 1/w = conjugate(w)
Then use the fact that a poly with real coefficients has roots occuring in conjugate pairs?

 

[Affinity]

Isn't it quicker for this equation to prove 1/w = conjugate(w)
Then use the fact that a poly with real coefficients has roots occuring in conjugate pairs?

The only problem is that 1/w = conjugate(w) only when |w| = 1.

So that doesn't work

sub in w and then sub in 1/w. For the 1/w equation times the whole thing by w^8 then ull get the same equation.

Minor detail: since (w =/= 0)

 

[MOP777]

The only problem is that 1/w = conjugate(w) only when |w| = 1.

So that doesn't work

Oh yeah.

...

I knew that, I was just testing that everyone else knew that.

 

[MetroMattums]

Nvm. Need to rethink this.

 

[Lukybear]

sub in w and then sub in 1/w. For the 1/w equation times the whole thing by w^8 then ull get the same equation.

How do u know w^8 =1?

 

[Lukybear]

P(z)=(z^4-5/4)^2-(3/4)^2
=(z^4-2)(z^4-1/2)

For this one, what do we do from here?

sorry for idiocy.

 

[Lukybear]

How do u know w^8 =1?

any1?

 

[shaon0]

For this one, what do we do from here?

sorry for idiocy.

Well, confirming 1/w is easy.
P(x)=(z^4-2)(z^4-1/2)
=(z^2-sqrt(2))(z^2+sqrt(2))(z^2-1/sqrt(2))(z^2+1/sqrt(2))
From here roots are:
z=+-2^(1/4),+-2^(-1/4),+-i.2^(1/4),+-i.2^(-1/4).

 

[Lukybear]

Sorry shaon0, could you just expand on how to confirm 1/w. Im thinking to do that question w^8 =1. But w is not a root of unity, but a root of p(z).

nd thxs for the roots bit. Ive got that.

 

[shaon0]

Sorry shaon0, could you just expand on how to confirm 1/w. Im thinking to do that question w^8 =1. But w is not a root of unity, but a root of p(z).

nd thxs for the roots bit. Ive got that.

Well, you've found complex roots;
Say, w=+-i(2)^(1/4), 1/w=1/(+-i.2^(1/4))=-+i.2^(-1/4) (as listed)
Try prod. of roots and sum.

 

[Lukybear]

Sorry still no clue.

Testing 1/w was a root was part a of the question. So...

 

[MetroMattums]

Sub 1/w in the equation, then take (1/w)^8 out of it, then you get the original equation, which is equal to 0. Hence, 1/w is a root.





For the first first first post, let x be a root of x^3 - 1 = 0 then let the roots of the new equation have a general form y = 1/(ax+b). Then, get those roots in terms of x then sub it into x^3 - 1 = 0. You should get the equation.