Squares in a fixed ratio with a linear offset: The square of one number is 224 less than 8 times the square of the other. If the numbers are in the ratio 3:4, find the two numbers.

Introduction / Context:
The problem pairs a ratio condition with a relation between squares. By parameterizing the numbers using the ratio and substituting into the square relation, we can solve for the scale factor and hence determine the actual numbers.

Given Data / Assumptions:

• Numbers are in the ratio 3:4 → let them be 3k and 4k
• The square of one is 224 less than 8 times the square of the other

Concept / Approach:
Test which arrangement satisfies the condition. Assume (4k)^2 = 8*(3k)^2 − 224 and solve for k. If it yields a consistent positive k, we have the valid assignment. Otherwise, try the converse.

Step-by-Step Solution:
Assume b = 4k, a = 3k and b^2 = 8a^2 − 22416k^2 = 8 * 9k^2 − 224 = 72k^2 − 224Rearrange: 16k^2 − 72k^2 = −224 → −56k^2 = −224 → k^2 = 4 → k = 2Numbers = 3k, 4k = 6, 8

Verification / Alternative check:
Check the given relation: 8*(6^2) − 224 = 8*36 − 224 = 288 − 224 = 64, and 8^2 = 64, so it holds exactly.

Why Other Options Are Wrong:
3,6 and 4,8 do not satisfy the exact offset of 224 under the 8x square relation. 6,4 flips the ratio order. 9,12 keeps 3:4 but fails the square condition.

Common Pitfalls:
Interpreting “224 less than” in the wrong direction. Always place the subtraction from the larger expression as stated.

Final Answer:
6 , 8