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October 14th, 2018, 03:19 PM  #1 
Member Joined: Sep 2011 Posts: 99 Thanks: 1  Proof using Rolle's Theorem
Hi, I have done up the proof for the question below. Please correct me if I have done wrong for the proof. Thanks in advanced! Question: Prove that if ab < 0 then the equation ax^3 + bx + c = 0 has at most three real roots. Proof: Let f(x) = ax^3 + bx + c. Assume that f(x) has 4 distinct roots, f(p) = f(q) = f(r) = f(s) = 0, there is a point x1 \in (p,q) such that f'(x1) = 0 ; x2 \in (q, r) such that f'(x2) = 0 ; x3 \in (r,s) such that f'(x3) = 0. Since ab < 0 then there are two possibilities where a>0 and b<0 or a <0 , b > 0. f'(x) = 3ax^2+b If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 > 0 and b < 0, then f'(x) = 0 If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 < 0 and b > 0, then f'(x) = 0 This is not true because the equation f'(x) = 0 has only two roots. Hence the given equation has at most three real roots when ab < 0. 
October 15th, 2018, 02:52 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,388 Thanks: 2015 
Every cubic equation has at most three real roots.

October 15th, 2018, 07:14 PM  #3 
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  That's what I was thinking too when I saw the question. However, it needs to be proven (for practice)
Last edited by ProofOfALifetime; October 15th, 2018 at 07:22 PM. 
October 15th, 2018, 07:22 PM  #4  
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Quote:
Honestly, the steps that come after need some work. I'm super tired right now, but I'd be happy to look at this more tomorrow if someone else doesn't respond.  

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