Question

three consecutive odd numbers are so such thst the square of first two is greater than the square of third by 65. find the numbers​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

11, 13, 15

Step-by-step explanation:

Let the first odd number be x

The other two conservative odd numbers will be (x + 2) and (x + 4)

Sum of the square of first two numbers is greater than the third by 65

⇒ x² + (x + 2)² = (x + 4)² + 65

Solve x:

x² + (x + 2)² = (x + 4)² + 65

x² + x² + 4x + 4 = x² + 8x + 16 + 65

2x² + 4x + 4 = x² + 8x + 81

x² - 4x - 77 = 0

(x + 7) (x - 11) = 0

x = 11 or x = -7 (rejected, assumed only positive number)

Find the numbers:

first number = x = 11

second number = x + 2 = 11 + 2 = 13

third number = x + 4 = 11 + 4 = 15

Answer: The numbers are 11, 13 and 15