Try to solve a quartic equation
in [Math]
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From: Wood on 30 Jul 2010 23:53 I am trying to learn how to solve a quartic equation. In "Summary of Ferrari's method" of page http://en.wikipedia.org/wiki/Quartic_function a root in the most simple case is x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) My question is how to know the express inside sqrt is not negative: how to know alfa^2 - 4*gama >= 0 and (-alfa + sqrt(alfa^2 - 4*gama))/2 >= 0
From: José Carlos Santos on 31 Jul 2010 04:28 On 31-07-2010 8:53, Wood wrote: > I am trying to learn how to solve a quartic equation. In > "Summary of Ferrari's method" of page > http://en.wikipedia.org/wiki/Quartic_function > a root in the most simple case is > x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) > > My question is how to know the express inside sqrt is not negative: > how to know > alfa^2 - 4*gama>= 0 > and > (-alfa + sqrt(alfa^2 - 4*gama))/2>= 0 You don't know. Some times it is and some times it isn't. Best regards, Jose Carlos Santos
From: Wood on 31 Jul 2010 01:13 How to calculate a negative value inside sqrt like sqrt(-4) if you say sqrt(-4) = 2i, then 2 + sqrt(-4) is a complex number and sqrt(2 + sqrt(-4)) will have two values and the quartic equation will have more than 4 roots!
From: Wood on 31 Jul 2010 01:20 Acording to the page, when beta = 0, the root is x = -B/4/A +-s sqrt((-alfa +-t sqrt(alfa^2 - 4*gama))/2) I think that is 4 roots: x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) x = -B/4/A + sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2) x = -B/4/A - sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) x = -B/4/A - sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2) Hence, I quess the expression under sqrt is a real number and is not negarive, and sqrt(expression) will be a single value.
From: José Carlos Santos on 31 Jul 2010 07:02
On 31-07-2010 10:20, Wood wrote: > Acording to the page, when beta = 0, the root is > x = -B/4/A +-s sqrt((-alfa +-t sqrt(alfa^2 - 4*gama))/2) > I think that is 4 roots: > x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) > x = -B/4/A + sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2) > x = -B/4/A - sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2) > x = -B/4/A - sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2) > > Hence, I quess the expression under sqrt is a real number and is not negarive, and sqrt(expression) will be a single value. Suppose that A = C = 1, that B = D = 0, and that E = 5/4. Then alpha = 1 and gamma = 5/4. Therefore, alpha^2 - 4*gamma = -4, which is negative. The square roots of alpha^2 - 4*gamma are 2i and -2i. Best regards, Jose Carlos Santos |