Roots of a Polynomial Function A²+B²+18C>0

In summary: Thank you for pointing that out.In summary, the given polynomial $P(x)=x^3+Ax^2+Bx+C$ with at least two distinct real roots, can be expressed in terms of its roots as $A=-(a+b+c)$, $B = bc+ca+ab$, and $C=-abc$. To prove that $A^2+B^2+18C>0$, the AM-GM inequality is used to show that $(a+b+c)^2+(bc+ca+ab)^2+18abc>0$. However, this may not hold true if two of the roots are negative, leading to a counterexample. Therefore, an extra condition is needed to exclude the case where two of
  • #1

anemone

Gold Member
MHB
POTW Director
3,883
115
If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.
 
Mathematics news on Phys.org
  • #2
anemone said:
If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.
[sp]Let $a,b,c$ be the three (real) roots of $P(x)$. Then $A = -(a+b+c)$, $B = bc+ca+ab$ and $C = -abc$. So we want to prove that $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 0.$$

Let $m = (abc)^{1/3}$ be the geometric mean of $a,b,c$. Since those numbers are not all equal, the AM-GM inequality is strict, so that $a+b+c > 3m$. For the same reason, $bc+ca+ab > 3m^2$. Therefore $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 9m^2 + 9m^4 + 18m^3 = 9m^2(1-m)^2 \geqslant0.$$
[/sp]
 
  • #3
Opalg said:
[sp]Let $a,b,c$ be the three (real) roots of $P(x)$. Then $A = -(a+b+c)$, $B = bc+ca+ab$ and $C = -abc$. So we want to prove that $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 0.$$

Let $m = (abc)^{1/3}$ be the geometric mean of $a,b,c$. Since those numbers are not all equal, the AM-GM inequality is strict, so that $a+b+c > 3m$. For the same reason, $bc+ca+ab > 3m^2$. Therefore $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 9m^2 + 9m^4 + 18m^3 = 9m^2(1-m)^2 \geqslant0.$$
[/sp]

Hello Opalg

Cannot apply AM-GM inequality as a,b,c are not positive
 
  • #4
kaliprasad said:
Hello Opalg

Cannot apply AM-GM inequality as a,b,c are not positive
[sp]Good point – I completely overlooked that. However, if one or three of the roots are negative then $C$ will be positive, so the inequality $A^2+B^2+18C>0$ will certainly hold. So the remaining case to deal with is if two of the roots are negative and the third one is positive. I'll have to think about that ... .

[/sp]
Edit:
[sp]The polynomial $x^3 + x^2 - x - 1 = (x+1)^2(x-1)$ has $A=1$, $B=C=-1$, and $A^2 + B^2 + 18C = -16 <0$. So I think that the problem probably needed an extra condition to exclude the case where two of the roots are negative.

[/sp]
 
Last edited:
  • #5
Opalg said:
[sp]
Edit:
[sp]The polynomial $x^3 + x^2 - x - 1 = (x+1)^2(x-1)$ has $A=1$, $B=C=-1$, and $A^2 + B^2 + 18C = -16 <0$. So I think that the problem probably needed an extra condition to exclude the case where two of the roots are negative.

[/sp]

I just checked the source of the problem, I didn't leave out anything. But you made the point, Opalg, that one such counterexample is suffice to disprove the validity of the problem. The problem is only valid if the condition to exclude the case where two of the real roots are negative is in place.
 

Suggested for: Roots of a Polynomial Function A²+B²+18C>0

Replies
14
Views
351
Replies
23
Views
968
Replies
5
Views
809
Replies
1
Views
695
Replies
7
Views
1K
Replies
4
Views
1K
Replies
11
Views
205
Back
Top