AMC 10 · 2011 · #16
Grade 8 arithmeticWhich of the following is equal to 9−62+9+62?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find the value of $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$ and match it to one of the answer choices.
Givens: The expression is $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$; It is a sum of two square roots that are almost the same, one with $-6\sqrt{2}$ inside and one with $+6\sqrt{2}$ inside; Answer choices: (A) $3\sqrt{2}$, (B) $2\sqrt{6}$, (C) $\frac{7\sqrt{2}}{2}$, (D) $3\sqrt{3}$, (E) $6$
Unknowns: The single simplified value of the whole expression
Understand
Restated: Find the value of $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$ and match it to one of the answer choices.
Givens: The expression is $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$; It is a sum of two square roots that are almost the same, one with $-6\sqrt{2}$ inside and one with $+6\sqrt{2}$ inside; Answer choices: (A) $3\sqrt{2}$, (B) $2\sqrt{6}$, (C) $\frac{7\sqrt{2}}{2}$, (D) $3\sqrt{3}$, (E) $6$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #5 Look for a Pattern, #3 Eliminate Possibilities
A messy nested-radical sum is hard to attack head-on, but a square root becomes easy the moment you square it. Tool #4 (Introduce a Variable) names the whole expression $S$ so we can square it once and turn two ugly roots into ordinary numbers. Squaring a sum leaves one leftover cross term, and Tool #7 (Identify Subproblems) isolates that cross term as its own product, which collapses by the difference-of-squares pattern. Tool #5 (Look for a Pattern) gives a clean second route: each radicand is secretly a perfect square, so the roots come out directly. Tool #3 (Eliminate Possibilities) is the safety net, since a quick decimal estimate separates the five choices, which are close but distinct.
Execute — Answer: B
8.EE.A.2 Step 1 Name the sum and square it
- Let $S$ be the whole expression.
- Both roots are positive, so $S>0$.
- Squaring a sum uses $(p+q)^2=p^2+q^2+2pq$, where here $p=\sqrt{9-6\sqrt{2}}$ and $q=\sqrt{9+6\sqrt{2}}$.
- Squaring each root just removes it, giving back the number inside.
💡 Squaring is the natural undo button for a square root, so squaring the sum trades roots for plain numbers.
7.NS.A.3 Step 2 Add the two plain terms
- The first two pieces are $9-6\sqrt{2}$ and $9+6\sqrt{2}$.
- The $-6\sqrt{2}$ and $+6\sqrt{2}$ cancel, leaving $9+9=18$.
💡 The two radicands are mirror images, so their irrational parts wipe each other out.
7.EE.A.1 Step 3 Collapse the cross term
- The product under the remaining root has the form $(a-b)(a+b)=a^2-b^2$, with $a=9$ and $b=6\sqrt{2}$.
- So it equals $9^2-\left(6\sqrt{2}\right)^2=81-72=9$.
- That leaves a clean $\sqrt{9}=3$ inside the cross term.
💡 A difference times a sum of the same two things always erases the middle and leaves $a^2-b^2$.
8.EE.A.2 Step 4 Take the positive square root
- Now $S^2=18+6=24$.
- Since $S$ is positive, $S=\sqrt{24}$.
- Pull out the perfect square: $\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}$.
- That is choice (B).
💡 Splitting off the largest perfect-square factor is how any square root gets to simplest form.
8.EE.A.2 Let $S$ be the whole expression. Both roots are positive, so $S>0$. Squaring a s 7.NS.A.3 The first two pieces are $9-6\sqrt{2}$ and $9+6\sqrt{2}$. The $-6\sqrt{2}$ and $ 7.EE.A.1 The product under the remaining root has the form $(a-b)(a+b)=a^2-b^2$, with $a= 8.EE.A.2 Now $S^2=18+6=24$. Since $S$ is positive, $S=\sqrt{24}$. Pull out the perfect sq Review
Reasonableness: Estimate with decimals: $6\sqrt{2}\approx 8.49$, so $\sqrt{9-8.49}=\sqrt{0.51}\approx 0.72$ and $\sqrt{9+8.49}=\sqrt{17.49}\approx 4.18$. Their sum is about $4.90$. Meanwhile $2\sqrt{6}\approx 2(2.449)=4.90$, an exact match. The nearby choices land elsewhere: $3\sqrt{2}\approx 4.24$, $\frac{7\sqrt{2}}{2}\approx 4.95$, $3\sqrt{3}\approx 5.20$, $6$. Only (B) hits $4.90$.
Alternative: Spot the hidden perfect squares (Tool #5). Since $\left(\sqrt{6}\mp\sqrt{3}\right)^2=6+3\mp 2\sqrt{18}=9\mp 6\sqrt{2}$, we get $\sqrt{9-6\sqrt{2}}=\sqrt{6}-\sqrt{3}$ and $\sqrt{9+6\sqrt{2}}=\sqrt{6}+\sqrt{3}$ (both positive because $\sqrt{6}>\sqrt{3}$). Adding them cancels the $\sqrt{3}$ terms and leaves $2\sqrt{6}$, again $\textbf{(B)}$.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions (Squaring the roots to remove them, and simplifying $\sqrt{24}=2\sqrt{6}$ by taking the positive root.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Adding $9-6\sqrt{2}$ and $9+6\sqrt{2}$ so the opposite irrational parts cancel to $18$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding the conjugate product $(9-6\sqrt{2})(9+6\sqrt{2})$ as the difference of squares $81-72=9$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Estimating each root as a decimal to confirm the sum is about $4.90$ and rule out the other choices.)
⭐ When a sum of square roots looks scary, name it and square it: the roots turn into plain numbers, and the leftover cross term usually collapses by difference of squares.
⭐ When a sum of square roots looks scary, name it and square it: the roots turn into plain numbers, and the leftover cross term usually collapses by difference of squares.
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