Divide 16 into two parts so that twice the square of the larger exceeds the square of the smaller by 164. Find the two parts (larger, smaller).

Introduction / Context:
This is a quadratic modeling question. Let the two parts be a (larger) and b (smaller) with a + b = 16. The condition links their squares: 2a^2 − b^2 = 164. Solve the system to identify the two numbers.

Given Data / Assumptions:

• a + b = 16 (a ≥ b ≥ 0 implicit)
• 2a^2 − b^2 = 164

Concept / Approach:
Express b as 16 − a and substitute into the square relation to obtain a single quadratic in a. Solve it, pick the physically meaningful root (nonnegative and larger), and then find b from the sum constraint.

Step-by-Step Solution:
Let b = 16 − aCondition: 2a^2 − (16 − a)^2 = 164Expand: 2a^2 − (256 − 32a + a^2) = 164 ⇒ a^2 + 32a − 256 = 164Simplify: a^2 + 32a − 420 = 0Discriminant: 32^2 + 4*420 = 1024 + 1680 = 2704 ⇒ √2704 = 52a = (−32 ± 52)/2 ⇒ a = 10 or a = −42 (reject)Thus a = 10; b = 16 − 10 = 6

Verification / Alternative check:
Check: 2*(10^2) − 6^2 = 200 − 36 = 164, satisfied. Sum 10 + 6 = 16, satisfied.

Why Other Options Are Wrong:
8,8 gives 2*64 − 64 = 64, not 164; 12,4 gives 288 − 16 = 272; 11,5 gives 242 − 25 = 217. Hence only 10,6 works.

Common Pitfalls:
Sign errors when expanding (16 − a)^2 or dropping the negative root without checking feasibility. Always verify both constraints.

Final Answer:
10, 6