AMC 10 · 2018 · #10

Grade 8 arithmetic

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

difference-of-squaresfraction-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: difference-of-squares

📏 Medium solution 💡 2 insights

Problem

A real number x makes the difference of two square roots equal 3: √(49-x²)-√(25-x²)=3. Find the value of the sum of those same two square roots, √(49-x²)+√(25-x²), and match it to one of the five choices.

Pick an answer.

(A)

(B)

$\sqrt{33}+8$

(C)

(D)

$2\sqrt{10}+4$

(E)

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable): name the two radicals a=√(49-x²) and b=√(25-x²) so the messy roots become two clean letters. The question gives a-b and asks for a+b — a difference-of-squares pairing. Tool #16 (Change Focus): instead of solving for x, focus on a²-b², because squaring the roots makes the x² cancel and turns the problem into pure arithmetic. Tool #7 (Identify Subproblems) splits the work into 'find a²-b²' then 'use (a-b)(a+b)=a²-b².' Tool #3 (Eliminate Possibilities) matches the final number to the choices.

1STEP 1

Name the two roots

Name the roots a=√(49-x²) and b=√(25-x²); both are nonnegative, so the radical tangle becomes two clean letters.

a=√(49-x²), b=√(25-x²)

2STEP 2

Restate given and goal

In these letters the given is a-b=3 and the goal is a+b — a known difference paired with an unknown sum.

a-b=3, find a+b

3STEP 3

Square each root

Squaring undoes each root: a²=49-x² and b²=25-x², leaving just the radicands behind.

a²=49-x², b²=25-x²

4STEP 4

Subtract to kill the x²

Subtracting the squared equations cancels the shared -x², leaving a²-b²=49-25=24.

a²-b²=(49-x²)-(25-x²)=24

5STEP 5

Use difference of squares

Difference of squares factors it as a²-b²=(a-b)(a+b), so (a-b)(a+b)=24.

a²-b²=(a-b)(a+b)=24

6STEP 6

Solve for the sum

Put a-b=3 into (a-b)(a+b)=24 to get 3(a+b)=24, so a+b=8 — choice (A).

3 (a+b)=24 → a+b=24/3=8 → (A)

Answer

The sum a+b must be larger than the difference a-b=3, and 8 ＞ 3, so the size is sensible. A quick consistency check: if a+b=8 and a-b=3, then a=5.5 and b=2.5, giving a²-b²=30.25-6.25=24, exactly matching the 49-25=24 from the radicands. The matching value x²=49-a²=49-30.25=18.75 is positive, so a real x truly exists. Everything is consistent.

💡Key takeaway

When you know the difference of two square roots and want their sum, multiply the pair: (a-b)(a+b)=a²-b², and here a²-b²=49-25=24, so the sum is 24÷3=8, choice (A).

Name the two roots
Restate given and goal
Square each root
Subtract to kill the x²
Use difference of squares
Solve for the sum