Nice solutions. Other approach based on Vietas formulas: let r, r and p be the roots of the cubic - we have the system of 3 equations: (1) 2r+p = 1/a (2) r^2+2rp = -1/a (3) r^2p = -2/a Solving this system for r we get the quadratic r^2+2r-6=0 just like in the video.
Nice solutions.
Other approach based on Vietas formulas: let r, r and p be the roots of the cubic - we have the system of 3 equations:
(1) 2r+p = 1/a
(2) r^2+2rp = -1/a
(3) r^2p = -2/a
Solving this system for r we get the quadratic r^2+2r-6=0 just like in the video.
Nice problem with ugly result