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
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If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.
 
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  • #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.
 

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

1. What is a polynomial function?

A polynomial function is a mathematical function that is expressed as the sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient. The highest power of the variable in a polynomial function is known as the degree of the function.

2. What are the roots of a polynomial function?

The roots of a polynomial function are the values of the variable that make the function equal to zero. In other words, they are the solutions to the equation formed by setting the polynomial function equal to zero.

3. How do you find the roots of a polynomial function?

The roots of a polynomial function can be found by factoring the function, setting each factor equal to zero, and solving for the variable. Alternatively, the roots can also be found by using the quadratic formula for second-degree polynomials or by using numerical methods for higher degree polynomials.

4. What does the inequality A²+B²+18C>0 represent for a polynomial function?

The inequality A²+B²+18C>0 represents the region in which the polynomial function has positive values. In other words, the values of the variable that satisfy this inequality will result in a positive output for the polynomial function.

5. How can the inequality A²+B²+18C>0 be used in real-world applications?

The inequality A²+B²+18C>0 can be used in real-world applications to determine the range of values for a variable that will result in a positive outcome. For example, it can be used in finance to determine the minimum amount of profit needed to break even or in physics to determine the range of values for a variable that will result in a positive velocity or acceleration.

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