Polynomial Q(s) (1 Viewer)

[pLuvia]

The equation px³+qx²+rx+s=0 has roots ac,a and a/c which are in geometric progression. Show that a=(-s/p)^1/3 and hence show that pr³-qs³=0

I got a=(-s/p)^1/3 but the next bit I got something very similar to it but just that the cubed is on the q not the s. Could someone confirm this for me please

Edit: My answer was pr³-q³s=0

 

[cyl123]

p(a)= p(-s/p)+q(-s/p)^2/3+r(-s/p)^1/3+s=0
-s + q(-s/p)^2/3+r(-s/p)^1/3+s=0
q(-s/p)^2/3 = -r(-s/p)^1/3
(q^3*s^2)/p^2 = -r^3(-s/p)
(q^3*s)/p = r^3
q^3*s= r^3*p
r^3*p - q^3*s =0

Unless I made an error, I think you are right

 

[speedie]

cyl u still online?

 

[vafa]

that question was from cambridge book. that is a mistake.
I emailed Arnold and he said that i am right.
Do not worry about that.

 

[pLuvia]

Ok thanks

 

[YBK]

Yup, the question is a mistake... or so does mrs arnold say


And your answer is right btw!