P(X) question (1 Viewer)

[bos1234]

The eqn

x^3 + 2x + 1 = 0 has roots a.b and c. FInd the equations with roots

a) a^2, B^2 and C^2

b) 1/a^2, 1/B^2 and 1/c^2

 

[Slidey]

The eqn

x^3 + 2x + 1 = 0 has roots a.b and c. FInd the equations with roots

a) a^2, B^2 and C^2

b) 1/a^2, 1/B^2 and 1/c^2

Oh I love these ones!!!

let y=x^2 -> x=sqrty
Sub this in:
ysqrty + 2sqrty + 1 = 0
sqrty(y+2) = -1
y(y+2)^2 = 1
y^3 + 4y^2 + 4y - 1 = 0
Thus your equation is:
x^3 + 4x^2 + 4x - 1 = 0

It's similar for the second one, but use y=1/x^2 -> x=1/sqrty as the substitution

EDIT: I can't see any errors...

 

[bos1234]

ahh kk

one more please..

Solve z^5 -4z = 0 over the field of complex numbers

thanks for the help

 

[Slidey]

Sorry, I deleted that because I noticed an error. I'll have a look after work and find it. I just haven't done 4u for a year so I don't want to teach you the wrong thing.

I'm moving this to the 4u forum - both of these topics are pretty certainly 4u (the complex field is not introduced in 3u).

Oh, and do you know of roots of unity? If not, you can't really solve that problem easily.

z^5-4z=0
z(z^4-4)=0

Obviously a root is z=0.
Let's find the others - they are the roots of z^4=4=4cis(0+2kpi)
z_k=sqrt2.cis(kpi/2)

z₀=sqrt2.cis(0)=sqrt2
z₁=sqrt2.cis(pi/2) = sqrt2.i
z₂=sqrt2.cis(pi)=-sqrt2
z₃=sqrt2.cis(3pi/2)=-sqrt2.i

Doh! That could have been solved using basic factorisation:
z^4-4=(z^2+2)(z^2-2)=(z+sqrt2.i)(z-sqrt2.i)(z+sqrt2)(z-sqrt2).

 

[pLuvia]

These questions are pretty common in the HSC papers (for 4u, not 3u) so it's good to learn the method Slidey has shown you

 

[bos1234]

k i need to revisit unities again. I thoguht there might b another way to solve them

thanks and bye

 

[Slidey]

Re-read my post. That specific question is possibly easier to solve with factorisation... depending on how intuitive roots of unity are to you.

If you mean a general way other than roots of unity... possibly. I'm not aware of one, though.

Haha funny sig.