Question

the sum of square of two positive integer is 202 if the square of the larger number is 40 more than the smaller number find those number​

Answer 1

Answer:

The given numbers are 9 and 11.

Step-by-step explanation:

Let the two positive integers be α and β.

∵ α² + β² = 202

and  α² - β² = 40

Adding both equations

2α² = 242

⇒ α² = 121

⇒ α = 11 (since the integers are positive)

Therefore β = √(202 - 121) = √81 = 9

α = 11, β = 9

(Note: question needs minor correction: if the square of the larger number is 40 more than the square of the smaller number)