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January 16th, 2015, 03:20 PM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  Finding of the constant term for the function to have distinct root
Hi I was solving this question. The function f(x)=x^312x+k has three distinct roots if and only if k is strictly between a and b. Find ab Now in the solve they assumed that the function have 2 distinct roots and then they solve it and find the value of a and b (the range of k). Then they use this value as the final answer. How come two root is same as three root when the question clearly stated distinct root? Thanks and Regards 
January 17th, 2015, 05:57 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
If a cubic equation has all three roots "distinct" then it is certainly true that any two of them are distinct.

January 19th, 2015, 10:33 AM  #3 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi country boy The solution looked like this. They assumed that it has two distinct root p & q. so the equation should be (xp)*(xp)*(xq)=0 and given x^312x+k=0 so 2p+q=0, p^2+2pq=12 and p^2q=k. Buy solving first two we can get values of p &q and replacing them in equation 3 we will get two different values of k. Now the solution said as k is in between a (16 in this particular case) and b(16 specifically) so the difference is 32. Now the whole solve was done for two root. I'm confused in this portion. How two root solve is true for 3 root? Last edited by Shen; January 19th, 2015 at 10:34 AM. Reason: partially written message posted 

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