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Question
Let P1(x)=ax2−bx−c, P2(x)=bx2−cx−a, P3(x)=cx2−ax−b be three quadratic polynomials where a,b,c are non-zero real numbers. Suppose there exists a real number α such that P1(α)=P2(α)=P3(α). Prove that a=b=c.
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Solution
We have three relations:
aα2−bα−c=λ,
bα2−cα−c=λ,
cα2−aα−c=λ,
where λ is the common value. Eliminating α2 from these, taking these equations pair-wise,we get three relations:
(ca−b2)(bc−a2)=(b−a),(ab−c2)(ca−b2)=(c−b),(bc−a2)(ab−c2)=(a−c).
Adding these three, we get
(ab+bc+ca−a2−b2−c2)(α−1)=0.
(Alternatively, multiplying above relations respectively by bc,ca and ab, and adding also leads to this.) Thus either ab+bc+ca−a2−b2−c2=0 or α=1. In the first case
0=ab+bc+ca−a2−b2−c2=12((a−b)2+(b−c)2+(c−a)2)
shows that a=b=c. If α=1, then we obtain
abc=bca=cab,
∴a=b=c.
aα2−bα−c=λ,
bα2−cα−c=λ,
cα2−aα−c=λ,
where λ is the common value. Eliminating α2 from these, taking these equations pair-wise,we get three relations:
(ca−b2)(bc−a2)=(b−a),(ab−c2)(ca−b2)=(c−b),(bc−a2)(ab−c2)=(a−c).
Adding these three, we get
(ab+bc+ca−a2−b2−c2)(α−1)=0.
(Alternatively, multiplying above relations respectively by bc,ca and ab, and adding also leads to this.) Thus either ab+bc+ca−a2−b2−c2=0 or α=1. In the first case
0=ab+bc+ca−a2−b2−c2=12((a−b)2+(b−c)2+(c−a)2)
shows that a=b=c. If α=1, then we obtain
abc=bca=cab,
∴a=b=c.
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