Q. polynomial -roots (1 Viewer)

[marsenal]

The question is:
Let a,b,c be the roots of the equation

x³ + px + q = 0 where q doesn't equal zero

Find the equation who roots are:

a^-1(1-a), b^-1(1-b), c^-1(1-c)

ie. 1-a over a etc...

Let x= (y+1)^-1

therefore the new equation will be:

qx³ + x²(p+3q) + x(2p+3q) + p + q + 1 = 0

is that close/right?

 

[bobo123]

can you explain to me how you got the constant to be 4?

 

[marsenal]

Sorry I should have put q+1. I've changed it now though

 

[bobo123]

i have the constant to be p+q+1

oh well
probably my mistake

 

[KeypadSDM]

Solution:
For polynomial P(x) = x^3 + px + q
Roots are a, b, and c

For polynomial with roots: (1-a) * a^-1, (1-b) * b^-1, and (1-c) * c^-1. Use the transformation, x = (x + 1) ^ -1

.: 0 = (x + 1) ^ -3 + p(x + 1) ^ -1 + q
0 = 1 + p(x + 1)^2 + q(x + 1)^3 (Multiply by (x + 1)^3)
0 = 1 + p(x^2 + 2x + 1) + q(x^3 + 3x^2 + 3x + 1)
0 = qx^3 + (p + 3q)x^2 + (2p + 3q)x + (1 + p + q)

.: G(x) =qx^3 + (p + 3q)x^2 + (2p + 3q)x + (1 + p + q)
has roots (1-a) * a^-1, (1-b) * b^-1, and (1-c) * c^-1.
Where P(x) has roots, a, b, and c.

And it seems while i was writing that you corrected your original post, oh well.

 

[marsenal]

Yea that's what I got now. Stupid mistakes! Thanks!