Question

`x^3-px^2+qx-r=0` has non-zero roots `p`, `q` and `r`. `q=-r` and `p=-1/r`. Find `p`, `q` and `r`. Could anyone solve this? Thanks

Answer 1

`(p, q, r) = (-1, -1, 1)`

Explanation:

We have:

`0 = x^3-px^2+qx-r`

`color(white)(0) = (x-p)(x-q)(x-r)`

`color(white)(0) = x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr`

By equating coefficients we find:

`{ (p = p+q+r rarr q+r=0), (q = pq+qr+rp = 1-r^2-1 = -r^2 rarr r=1) :}`

So `p = -1/r = -1/1 = -1` and `q = -r = -1`