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January 5th, 2019, 08:46 AM  #1 
Newbie Joined: Jan 2019 From: Israel Posts: 2 Thanks: 0  Prove general solution for deppresed cubic equation y^3+py+q=0 given a discriminant
I've been given the following problem and I can't seem to figure out the solution, would appreciate any help and direction to solution ! Let's look at the equation : y3 + py + q = 0 (*) We shall define Delta (or d in short) d = 4p3 + 27q2 Prove that : a) If d > 0 , then (*) has a single solution . b) If d = 0 , and at least one of the coefficients (p , q) =/= 0, then (*) has 2 solutions . c) If d < 0 , then (*) has 3 solutions . Thank you very much in advance ! 
January 5th, 2019, 01:10 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
You could write the equation as $(ya)(yb)(yc)=0$ Compute p and q, noting that the coefficient of $y^2$ is 0 and then set up d. Then look at what happens to a, b, and c.

January 5th, 2019, 01:16 PM  #3 
Newbie Joined: Jan 2019 From: Israel Posts: 2 Thanks: 0 
Thank you very much !


Tags 
cubic, deppresed, discriminant, equation, general, prove, solution 
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