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Question:

What is the maximum number of real roots for this equation? Assume $a,b,c$ are positive real numbers. $$(ax^2+bx+c)(bx^2+cx+a)(cx^2+ax+b)=0$$

My attempt:

Let us assume the equation has 6 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc>0$.
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a>2\sqrt{bc}$
Multiply all of them to get $abc>8abc$ which is a contradition for positive real numbers.

Now let us assume the equation has 5 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc=0$. (modifying the last discrimnant here, but we could choose any other discriminant as well since all are equivalent)
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a=2\sqrt{bc}$
Multiply the first two to get $bc>4a\sqrt{bc}=>\sqrt{bc}>4a$ which is a contradition since $a=2\sqrt{bc}$.

Now let us assume the equation has 4 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc<0$. (modifying the last discrimnant here, but we could choose any other discriminant as well since all are equivalent)
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a<2\sqrt{bc}$
Multiply the first two to get $bc>4a\sqrt{bc}\implies a<\sqrt{bc}/4$ which is possible for a particular combination of $a,b,c$ like $a=1,b=16,c=27$

I attempted this question on my own (self-study) and wish to know if my method is correct as I usually miss important assumptions and related things.

I hope the question is clear enough. In case of any ambiguity please comment.

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Gaurang Tandon
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  • You still need to check if any of the roots are the same. – DHMO Apr 16 '17 at 12:25
  • @DHMO The OP has shown that it is impossible that the equation has more than $4$ roots and that it can have $4$ roots. I don't see the need of messing up with repeated roots. – ajotatxe Apr 16 '17 at 12:30
  • @ajotatxe For the last example, the OP managed to construct an example. However, the OP has not checked if the equation actually has $4$ roots (although I checked graphically that it does indeed have $4$ roots). – DHMO Apr 16 '17 at 12:30
  • @DHMO Yes, sorry, you are right. – ajotatxe Apr 16 '17 at 12:31
  • @DHMO Yep, good catch. Thanks! – Gaurang Tandon Apr 16 '17 at 15:57

2 Answers2

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If find it fairly good. There is only a little mistake (for copy-pasting, probably) in the case of $4$ real roots, when you say "multiply all of them", but further reading makes clear what you mean.

Another detail. If the equation has $4$ real roots, another possibility is that two discriminants are zero and the other is positive, but this leads to contradiction too. Perhaps you can get rid of zero discriminants at once if you assume at first that all the discriminants are non negative, instead of strictly positive. You get $abc\ge 8abc$ which is also a contradiction.

EDIT: See also DHMO's comment ("OP managed to construct an example. However, the OP has not checked if the equation actually has 4 roots.")

Gaurang Tandon
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ajotatxe
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  • "if you assume at first that all the discriminants are non negative" -> yep, that's good. That way I would be invalidating the different invalid ways of getting 4,3 roots as well - at once - 6 real roots (all $\Delta$ > 0), 5 real roots (one $\Delta$ = 0, two $\Delta$ > 0), 4 real roots (two $\Delta$ = 0, one $\Delta$ > 0), 3 real roots (when all one $\Delta$ = 0). – Gaurang Tandon Apr 16 '17 at 16:01
  • I edited my question to remove that typo which you're talking about in the first line. – Gaurang Tandon Apr 17 '17 at 10:16
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The assumption that a, b and c are 'positive' numbers limits the range of answers. Once you assume that a, b and c are just real (positive or negative) numbers, the answer to your question is 6 real roots. This maximum number of 6 real roots is obtained when for instance b is positive and c is negative or vice versa. You can check it with this example: a equals 1, b equals minus 5 and c equals 6.

Flatortue Method
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