AMC 10 · 2021 · #2

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition

absolute-valuesigned-square-rootsign-analysisexponents identify-subproblemscasework ↑ Prerequisites: absolute-valueexponents

📏 Short solution 💡 2 insights

Problem

What is the value of $\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}$ ?

Pick an answer.

(A)

(B)

$~4\sqrt{3}-6$

(C)

(D)

$~4\sqrt{3}$

(E)

$~4\sqrt{3}+6$

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Toolkit + CCSS Solution

Understand

Restated: Evaluate $\sqrt{(3-2\sqrt{3})^2} + \sqrt{(3+2\sqrt{3})^2}$, using the fact that $\sqrt{a^2}$ equals $|a|$ for any real $a$.

Givens: The expression $\sqrt{(3-2\sqrt{3})^2} + \sqrt{(3+2\sqrt{3})^2}$; $\sqrt{a^2} = |a|$ (the principal square root is non-negative); Answer choices: (A) $0$, (B) $4\sqrt{3} - 6$, (C) $6$, (D) $4\sqrt{3}$, (E) $4\sqrt{3} + 6$

Unknowns: The numerical value of the expression

Understand

Restated: Evaluate $\sqrt{(3-2\sqrt{3})^2} + \sqrt{(3+2\sqrt{3})^2}$, using the fact that $\sqrt{a^2}$ equals $|a|$ for any real $a$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #9 (Easier Problem) — replace the messy expression $\sqrt{(\square)^2}$ with the cleaner rule $|\square|$. That turns the problem into a sum of two absolute values, which is concrete. Tool #7 (Subproblems) splits the work into "sign of $3 - 2\sqrt{3}$" and "sign of $3 + 2\sqrt{3}$" — each handled separately, then added. Tool #3 (Eliminate) gives a fast sanity check: $2\sqrt{3} \approx 3.46$, so $3 - 2\sqrt{3} \approx -0.46$ and $3 + 2\sqrt{3} \approx 6.46$; their absolute values sum to $\approx 6.92 \approx 4\sqrt{3}$. Only (D) matches.

Execute — Answer: D

#9 Solve an Easier Related Problem 8.EE.A.2 Step 1

Replace each square-root-of-a-square with an absolute value.
For any real number $a$, $\sqrt{a^2} = |a|$ — the principal square root is never negative.

$$\sqrt{(3-2\sqrt{3})^2} + \sqrt{(3+2\sqrt{3})^2} = |3-2\sqrt{3}| + |3+2\sqrt{3}|$$

💡 Squaring then square-rooting just throws away the sign — Grade 8 square-root standard.

#7 Identify Subproblems 8.NS.A.2 Step 2

Compare $3$ and $2\sqrt{3}$ to decide each sign.
Square both: $3^2 = 9$ and $(2\sqrt{3})^2 = 4 \cdot 3 = 12$.
Since $12 > 9$ and both are positive, $2\sqrt{3} > 3$.

$$3^2 = 9 < 12 = (2\sqrt{3})^2 \;\Rightarrow\; 2\sqrt{3} > 3$$

💡 Comparing an irrational to a rational by squaring — Grade 8 "rational approximations of irrationals".

#7 Identify Subproblems 7.NS.A.1 Step 3

Peel each absolute value.
Since $2\sqrt{3} > 3$, $3 - 2\sqrt{3} < 0$, so $|3 - 2\sqrt{3}| = -(3 - 2\sqrt{3}) = 2\sqrt{3} - 3$.
And $3 + 2\sqrt{3} > 0$, so $|3 + 2\sqrt{3}| = 3 + 2\sqrt{3}$.

$$|3-2\sqrt{3}| = 2\sqrt{3} - 3, \quad |3+2\sqrt{3}| = 3 + 2\sqrt{3}$$

💡 Negate to make non-negative — Grade 7 absolute value of a difference.

#7 Identify Subproblems 7.EE.A.1 Step 4

Add the two pieces and combine like terms.
The $-3$ and $+3$ cancel, leaving two copies of $2\sqrt{3}$, which is $4\sqrt{3}$.
That is choice (D).

$$(2\sqrt{3} - 3) + (3 + 2\sqrt{3}) = 4\sqrt{3} \;\Rightarrow\; \textbf{(D)}$$

💡 Combine like terms (rational $+$ irrational separately) — Grade 7 linear-expression standard.

#3 Eliminate Possibilities 8.NS.A.2 Step 5

Sanity check with decimals.
$2\sqrt{3} \approx 3.46$, so the two pieces are about $0.46$ and $6.46$.
Their sum $\approx 6.92$, and $4\sqrt{3} \approx 6.93$ — matches (D) exactly, and no other choice fits.

$$0.46 + 6.46 \approx 6.92 \approx 4\sqrt{3}$$

💡 Decimal approximation confirms the symbolic answer — Grade 8 irrational approximations.

[1] #9 8.EE.A.2 Replace each square-root-of-a-square with an absolute value. For any real number

[2] #7 8.NS.A.2 Compare $3$ and $2\sqrt{3}$ to decide each sign. Square both: $3^2 = 9$ and $(2\

[3] #7 7.NS.A.1 Peel each absolute value. Since $2\sqrt{3} > 3$, $3 - 2\sqrt{3} < 0$, so $|3 - 2

[4] #7 7.EE.A.1 Add the two pieces and combine like terms. The $-3$ and $+3$ cancel, leaving two

[5] #3 8.NS.A.2 Sanity check with decimals. $2\sqrt{3} \approx 3.46$, so the two pieces are abou

Review

Reasonableness: Both terms inside the square roots are real, so each square root is well-defined and non-negative. The sum $4\sqrt{3} \approx 6.93$ is positive and lies between (C) $6$ and (E) $4\sqrt{3} + 6 \approx 12.93$, exactly where the rough estimate places it. The $-3$ and $+3$ cancellation is the key structural fingerprint of choice (D).

Alternative: Tool #6 (Guess and Check) — pick numerical values: $3 \approx 3$, $2\sqrt{3} \approx 3.464$. Each square root is $\sqrt{0.464^2} \approx 0.464$ and $\sqrt{6.464^2} \approx 6.464$. Sum $\approx 6.928$, which matches $4\sqrt{3} \approx 6.928$. Same answer (D) without symbolic manipulation.

CCSS standards used (min grade 8)

7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Reading $|3 - 2\sqrt{3}|$ as a non-negative distance and flipping the sign of the negative inside.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Combining like terms $(2\sqrt{3} - 3) + (3 + 2\sqrt{3}) = 4\sqrt{3}$.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Comparing $3$ to $2\sqrt{3}$ by squaring and approximating $2\sqrt{3} \approx 3.46$ to confirm the answer.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Applying $\sqrt{a^2} = |a|$ to each term to remove the square-root-of-a-square structure.)

⭐ This AMC 10 problem only needs Grade 8 "$\sqrt{a^2} = |a|$" — once you peel each square root into an absolute value and notice $2\sqrt{3} > 3$, the $-3$ and $+3$ cancel, leaving $4\sqrt{3}$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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