AMC 8 · 2019 · #20

Grade 6 algebra

Archetype Branching State Enumeration

perfect-squaressystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: perfect-squaresexponents

📏 Medium solution 💡 2 insights

Problem

How many different real numbers $x$ satisfy the equation $(x^{2}-5)^{2}=16?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Count how many different real numbers $x$ make the equation $(x^2 - 5)^2 = 16$ true. The answer must be one of $0, 1, 2, 4,$ or $8$.

Givens: Equation: $(x^2 - 5)^2 = 16$; The unknown $x$ is a real number; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $4$, (E) $8$

Unknowns: The number of distinct real values of $x$ that satisfy the equation

Understand

Restated: Count how many different real numbers $x$ make the equation $(x^2 - 5)^2 = 16$ true. The answer must be one of $0, 1, 2, 4,$ or $8$.

Givens: Equation: $(x^2 - 5)^2 = 16$; The unknown $x$ is a real number; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $4$, (E) $8$

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities, #2 Make a Systematic List, #7 Identify Subproblems

Whenever we see $(\text{something})^2 = 16$, 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 $x^2 - 5 = 4$?" and "when is $x^2 - 5 = -4$?" Tool #6 (Guess and Check) lets us scan small integers $x = 0, \pm 1, \pm 2, \pm 3, \pm 4$ 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.

Execute — Answer: D

#3 Eliminate Possibilities 6.EE.A.1 Step 1

Translate the outer square.
The equation $(x^2 - 5)^2 = 16$ is true exactly when $x^2 - 5$ is one of the two real numbers whose square is $16$, namely $+4$ or $-4$.

$$(x^2 - 5)^2 = 16 \;\Longleftrightarrow\; x^2 - 5 = 4 \;\text{or}\; x^2 - 5 = -4$$

💡 Knowing that only $4$ and $-4$ square to $16$ uses the meaning of the exponent $2$ — a Grade 6 idea.

#7 Identify Subproblems 6.EE.B.7 Step 2

Rewrite each case as a simple equation about $x^2$.
Add $5$ to both sides of each branch.

Case 1: $x^2 = 9. \qquad$ Case 2: $x^2 = 1.$

💡 Splitting one equation into two easier ones is the Grade 6 "solve an equation of the form $p\,x = q$" pattern.

#6 Guess and Check 6.NS.C.6 Step 3

Hunt for integers that satisfy $x^2 = 9$ by guessing small whole numbers and their negatives.
List the candidates in order.

$0^2 = 0,\; (\pm 1)^2 = 1,\; (\pm 2)^2 = 4,\; (\pm 3)^2 = 9.\;$ Hits: $x = 3$ and $x = -3$.

💡 Trying both $+3$ and $-3$ on a number line is exactly what Grade 6 "positive and negative numbers" expects.

#6 Guess and Check 6.NS.C.6 Step 4

Do the same scan for $x^2 = 1$.
Continue from the table above.

From the same table, $(\pm 1)^2 = 1.\;$ Hits: $x = 1$ and $x = -1$.

💡 Reusing the squared-integer table avoids extra work and reinforces that $(-1)^2 = 1$ just like $1^2 = 1$.

#2 Make a Systematic List 6.EE.B.5 Step 5

Confirm no other real $x$ works.
For $|x| \geq 4$, the inside $x^2 - 5 \geq 11$, so $(x^2-5)^2 \geq 121 > 16$; the value only grows from there.
And between consecutive integers, $x^2$ moves smoothly from one square to the next, so no extra integer hit appears outside our list.
Combine the two cases into one systematic list.

All solutions: $\{-3,\, -1,\, 1,\, 3\}.\;$ Count $= 4 \;\Rightarrow\; \textbf{(D)}.$

💡 Listing every value that satisfies the equation in order is the Grade 6 "find all values that make an equation true" idea.

[1] #3 6.EE.A.1 Translate the outer square. The equation $(x^2 - 5)^2 = 16$ is true exactly when

[2] #7 6.EE.B.7 Rewrite each case as a simple equation about $x^2$. Add $5$ to both sides of eac

[3] #6 6.NS.C.6 Hunt for integers that satisfy $x^2 = 9$ by guessing small whole numbers and the

[4] #6 6.NS.C.6 Do the same scan for $x^2 = 1$. Continue from the table above.

[5] #2 6.EE.B.5 Confirm no other real $x$ works. For $|x| \geq 4$, the inside $x^2 - 5 \geq 11$,

Review

Reasonableness: Quickly verify each solution by plugging back in: $((\pm 3)^2 - 5)^2 = (9-5)^2 = 16 \checkmark$ and $((\pm 1)^2 - 5)^2 = (1-5)^2 = (-4)^2 = 16 \checkmark$. 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 $(x^2-5)^2$ keeps growing for $|x| > 3$, so no more exist. Choice (D) is consistent.

Alternative: Tool #13 (Convert to Algebra) via substitution: let $u = x^2 - 5$. Then $u^2 = 16 \Rightarrow u = \pm 4$, giving the same two cases. Each case is a quadratic in $x$ with two real roots, totaling $4$ solutions. The substitution dresses up the same logic but skips the kid-friendly Guess-and-Check feel.

CCSS standards used (min grade 6)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Recognizing that $(x^2 - 5)^2 = 16$ forces $x^2 - 5 = \pm 4$ because only $\pm 4$ square to $16$, and evaluating $x^2$ for small integers in the scan.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Rewriting each branch as the simpler equation $x^2 = 9$ or $x^2 = 1$ by adding $5$ to both sides.)
6.NS.C.6 Understand a rational number as a point on the number line (Recognizing that both $+3$ and $-3$ (and both $+1$ and $-1$) are distinct points whose squares are equal.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Combining the hits from both cases into one list and counting all real values of $x$ that make the original equation true.)

⭐ This AMC 8 problem only needs Grade 6 exponents and the rule that both $+a$ and $-a$ square to $a^2$ — concepts you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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