Indian Regional Mathematical Olympiad - 2007
Find the number of ordered pairs of real numbers such that whenever is a root , is also a root of the equation.
The answer is 6.
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1 solution
Consider the equation x2 + ax + b = 0. It has two roots (not necessarily real), say α and β.
Either α = β or α ≠ β.
Case-I:--
Suppose α = β, so that α is a double root. Since α2 – 2 is also a
root, the only possibility is α = α2 – 2. This
reduces to (α + 1) (α – 2) = 0. Hence α = –1 or α = 2. Observe
that a = –2α and b = α2. Thus (a, b) = (2, 1)
or (–4, 4).
Case-II:
Suppose α ≠ β. There are four possibilities
(I) α = α2 – 2 and β = β2 – 2;
(II) α = β2 – 2 and β = α2 – 2;
(III) α = α2 – 2 = β2 – 2 and α ≠ β; or
(IV) β = α2 – 2 = β2 – 2 and α ≠ β.
(a) Here (α, β) = (2, –1) or (–1, 2). Hence (a, b) = (– (α + β), αβ) = (–1, –2)
(b) Suppose α = β2 – 2 and β = α2 – 2. Then
α – β = β2 – α2 = (β – α) (β + α)
Since α ≠ β, we get β + α = –1. However, we also have
α + β = β2 + α2 – 4 = (α + β)2 – 2αβ – 4
Thus –1 = 1 – 2αβ – 4, which implies that αβ = –1. Therefore (a, b) = (– (α + β), αβ) = (1, –1).
(c) If α = α2 – 2 = β2 – 2 and α ≠ β, then α = – β. Thus α = 2, β = –2 or α = –1, β = 1.
In this case (a, b) = (0, –4) and (0, –1).
(d) Note that β = α2 – 2 = β2 – 2 and α ≠ β is identical to (III), so that we get exactly same pairs (a, b).
Thus we get 6 pairs; (a, b) = (–4, 4), (2, 1), (–1, –2), (1, –1), (0, –4), (0, –1).
SO ANSWER SHOULD BE 6