MHB Quadratic Polynomial Problem: Proving Equality with Real Numbers

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The problem involves three quadratic polynomials, P1, P2, and P3, defined with non-zero real coefficients a, b, and c. It is given that there exists a real number k such that P1(k) equals P2(k) and P3(k). The task is to prove that this equality implies a, b, and c must all be equal. A suggested solution is provided, although no responses were recorded for the previous problem of the week. The discussion highlights the challenge of proving the equality of coefficients in the context of polynomial functions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

Let $P_1(x)=ax^2-bx-c,\,P_2(x)=bx^2-cx-a,\,P_3(x)=cx^2-ax-b$ be three quadratic polynomials where $a,\,b$ and $c$ are non-zero real numbers. Suppose there exists a real number $k$ such that $P_1(k)=P_2(k)=P_3(k)$, prove that $a=b=c$.

-----

 
Physics news on Phys.org
No one answered last week's POTW.(Sadface) However, you can find the suggested solution below:

We have three relations:

$ak^2-bk-c=m\\ bk^2-ck-a=m\\ ck^2-ak-b=m$
where $m$ is the common value.

Eliminating $k^2$ from these, taking these equations pair-wise, we get:

$(ca-b^2)k-(bc-a^2)=m(b-a)\\(ab-c^2)k-(ca-b^2)=m(c-b)\\(bc-a^2)k-(ab-c^2)=m(a-c)$

Adding these three we get

$(ab+bc+ca-a^2-b^2-c^2)(k-1)=0$

Thus, either $ab+bc+ca-a^2-b^2-c^2=0$ or $k=1$.

In the first case

$ab+bc+ca-a^2-b^2-c^2=0\\ \dfrac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)=0$
shows that $a=b=c$.

If $k=1$, then we obtain $a-b-c=b-c-a=c-a-b$, once again we obtain $a=b=c$.
 
Back
Top