AMC 10 · 2003 · #25

Grade 9 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

domain-restrictionquadratic-equationscompleting-the-square caseworkguess-and-check ↑ Prerequisites: domain-restriction

📏 Long solution 💡 3 insights

Problem

For the function f(x) = √(ax² + bx), count the real numbers a for which at least one positive b makes the domain of f and the range of f be exactly the same set of numbers.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

How to solve

Strategy Eliminate Possibilities

The shape of f changes completely with the sign of a: for a ＞ 0 the radicand is an upward parabola, for a = 0 it is a line, for a ＜ 0 it is a downward parabola. So Tool #7 (Identify Subproblems) splits the question into those three cases, and Tool #3 (Eliminate Possibilities) settles each one — the counting is complete only when all three are decided, and a = 0 must be checked on its own because dividing by a is illegal there. Inside the hardest case, a ＜ 0, both the domain and the range turn out to be intervals starting at 0, so Tool #4 (Introduce a Variable) with s = √(-a) turns matching them into a short equation. Tool #6 (Guess and Check) then verifies each surviving value of a on a concrete b rather than trusting the algebra.

1STEP 1

Write down what domain and range mean

A square root needs a nonnegative inside and returns a nonnegative value, so the range sits in [0, ∞).

domain = {x : x(ax + b) ≥ 0}, range ⊆ [0, ∞)

2STEP 2

Case a ＞ 0: the domain leaks negative numbers

If the leading coefficient is positive the domain leaks a negative number, so it can never match.

a ＞ 0: domain = (-∞, -b/a] ∪ [0, ∞) ∋ -b/a ＜ 0

3STEP 3

Case a = 0: check it on its own

The zero case must be checked on its own, and there both sets are [0, ∞) — it works.

a = 0: f(x) = √(bx), domain = [0,∞), range = [0,∞)

4STEP 4

Case a ＜ 0: find both sets as intervals

If it is negative, both sets are closed intervals starting at zero, found by completing the square — two endpoints to match.

a ＜ 0: domain = [0, -b/a], range = [0, b/(2√(-a))]

5STEP 5

Match the two intervals

Equating the right endpoints cancels b and gives exactly a = -4.

-b/a = b/(2√(-a)) → 1/s² = 1/2s → s = 2 → a = -4

6STEP 6

Verify and count

Verifying leaves exactly the two values zero and negative four, so the count is 2, choice (C).

a = -4, b = 4: domain = range = [0,1]; {a} = {0, -4} → (C) 2

Answer

The count is 2, and both survivors were confirmed by direct substitution rather than by algebra alone: a = 0 with f(x) = √(bx) gives [0,∞) both ways, and a = -4 with b = 4 gives [0,1] both ways. The elimination side is just as concrete: for a ＞ 0 the domain provably contains the negative number -b/a, and no range of a square root ever does, so (E) infinitely many is impossible and so is anything above 2. One trap deserves naming. It is tempting to derive a single equation such as a(a + 4) = 0 and read off a = 0 and a = -4 together, but that equation is obtained by dividing by a and writing √(-a), steps that are meaningless at a = 0. The root a = 0 produced that way is an artifact; a = 0 is genuinely a solution, but only because the separate check on f(x) = √(bx) says so.

💡Key takeaway

Split by the sign of the leading coefficient, work out the domain and range as intervals in each case, and check the boundary case a = 0 on its own — dividing by a can never tell you anything about a = 0.

Write down what domain and range mean
Case a ＞ 0: the domain leaks negative numbers
Case a = 0: check it on its own
Case a ＜ 0: find both sets as intervals
Match the two intervals
Verify and count