AMC 8 · 2019 · #20

Grade 6 algebra
perfect-squaressystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: perfect-squaresexponents
📏 Medium solution 💡 2 insights
Problem
Count how many different real numbers x make the equation (x² - 5)² = 16 true. The answer must be one of 0, 1, 2, 4, or 8.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
4
(E)
8

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

Whenever we see (something)² = 16, the "something" is forced to be either +4 or -4 — the only two real numbers whose square is 16. That turns one scary quartic into two friendly questions: "when is x² - 5 = 4?" and "when is x² - 5 = -4?" Tool #6 (Guess and Check) lets us scan small integers x = 0, ± 1, ± 2, ± 3, ± 4 to spot every value where the inside hits 9 or 1. Tool #2 (Systematic List) keeps the scan organized so nothing is missed, and Tool #3 (Eliminate) confirms the count matches choice (D) and rules out the other choices.

1STEP 1

Undo the outer square: (x² - 5)² = 16 holds exactly when the inside x² - 5 is +4 or -4.

(x² - 5)² = 16 ⟺ x² - 5 = 4 or x² - 5 = -4
2STEP 2

Add 5 to both sides of each branch, turning them into x² = 9 and x² = 1.

Case 1: x² = 9. Case 2: x² = 1.
3STEP 3

Scan small integers and their negatives for x² = 9 — the hits are x = 3 and x = -3.

0² = 0, (± 1)² = 1, (± 2)² = 4, (± 3)² = 9. Hits: x = 3 and x = -3.
4STEP 4

Do the same scan for x² = 1 — the hits are x = 1 and x = -1.

From the same table, (± 1)² = 1. Hits: x = 1 and x = -1.
5STEP 5

Nothing with |x| ≥ 4 or hidden between integers works, so the complete list is {-3, -1, 1, 3}.

All solutions: {-3, -1, 1, 3}. Count = 4 → (D).
Answer
4
Quickly verify each solution by plugging back in: ((± 3)² - 5)² = (9-5)² = 16 ✓ and ((± 1)² - 5)² = (1-5)² = (-4)² = 16 ✓. All four values are different (one positive, one negative pair from each case), so the count is exactly 4. Choice (E) 8 would require eight different x-values; we found only four, and (x²-5)² keeps growing for |x| > 3, so no more exist. Choice (D) is consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 6 exponents and the rule that both +a and -a square to a² — concepts you already know!