AMC 10 · 2011 · #16

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

difference-of-squaressigned-square-rootexponents convert-to-algebra ↑ Prerequisites: exponents

📏 Medium solution 💡 2 insights

Problem

Two nested square roots are added together:

\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}

. The two radicands differ only in the sign of

6\sqrt{2}

, and both are positive, so the sum is a positive real number. Find which answer choice this sum equals.

Pick an answer.

(A)

$3\sqrt2$

(B)

$2\sqrt6$

(C)

$\frac{7\sqrt2}{2}$

(D)

$3\sqrt3$

(E)

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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.

1STEP 1

Name the sum and square it

Call the whole positive expression $S$ and square it with $(p+q)^2=p^2+q^2+2pq$ : each squared root gives back the number inside.

S²=(9-6√(2))+(9+6√(2))+2√((9-6√(2))(9+6√(2)))

2STEP 2

Add the two plain terms

In $(9-6\sqrt{2})+(9+6\sqrt{2})$ the two $6\sqrt{2}$ terms cancel, leaving $18$ .

(9-6√(2))+(9+6√(2))=18

3STEP 3

Collapse the cross term

The product under the leftover root is a difference of squares: $9^2-(6\sqrt{2})^2=81-72=9$ , so the cross term is $2\sqrt{9}=6$ .

(9-6√(2))(9+6√(2))=81-72=9, 2√(9)=6

4STEP 4

Take the positive square root

Now $S^2=18+6=24$ , and $S$ is positive, so $S=\sqrt{24}=\sqrt{4\cdot 6}$ , which is $2\sqrt{6}$ — choice (B).

S²=24 → S=√(24)=2√(6) → (B) 2√(6)

Answer

2√6

Estimate with decimals: 6√(2)≈ 8.49, so √(9-8.49)=√(0.51)≈ 0.72 and √(9+8.49)=√(17.49)≈ 4.18. Their sum is about 4.90. Meanwhile 2√(6)≈ 2(2.449)=4.90, an exact match. The nearby choices land elsewhere: 3√(2)≈ 4.24, 7√(2)/2≈ 4.95, 3√(3)≈ 5.20, 6. Only (B) hits 4.90.

💡Key takeaway

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.

Name the sum and square it
Add the two plain terms
Collapse the cross term
Take the positive square root

Name the sum and square it

Add the two plain terms

Collapse the cross term

Take the positive square root

More like this