<Dl> <Dd> (r 1 + r 2 + r 3 + r 4 = 0 (r 1 + r 2) (r 3 + r 4) = − α (r 1 + r 3) (r 2 + r 4) = − β (r 1 + r 4) (r 2 + r 3) = − γ . (\ displaystyle \ left \ ((\ begin (array) (l) r_ (1) + r_ (2) + r_ (3) + r_ (4) = 0 \ \ (r_ (1) + r_ (2)) (r_ (3) + r_ (4)) = - \ alpha \ \ (r_ (1) + r_ (3)) (r_ (2) + r_ (4)) = - \ beta \ \ (r_ (1) + r_ (4)) (r_ (2) + r_ (3)) = - \ gamma (\ text (.)) \ end (array)) \ right .) </Dd> </Dl> <Dd> (r 1 + r 2 + r 3 + r 4 = 0 (r 1 + r 2) (r 3 + r 4) = − α (r 1 + r 3) (r 2 + r 4) = − β (r 1 + r 4) (r 2 + r 3) = − γ . (\ displaystyle \ left \ ((\ begin (array) (l) r_ (1) + r_ (2) + r_ (3) + r_ (4) = 0 \ \ (r_ (1) + r_ (2)) (r_ (3) + r_ (4)) = - \ alpha \ \ (r_ (1) + r_ (3)) (r_ (2) + r_ (4)) = - \ beta \ \ (r_ (1) + r_ (4)) (r_ (2) + r_ (3)) = - \ gamma (\ text (.)) \ end (array)) \ right .) </Dd> <P> It is a consequence of the first two equations that r + r is a square root of α and that r + r is the other square root of α . For the same reason, </P> <Ul> <Li> r + r is a square root of β, </Li> <Li> r + r is the other square root of β, </Li> <Li> r + r is a square root of γ, </Li> <Li> r + r is the other square root of γ . </Li> </Ul>

How to find the solutions of a quartic equation