Question

Find two consecutive positive odd numbers such that the square of their sum exceeds the sum of their squares by 126

Answer 1

Answer :-

Let the number be x

Other number = x + 2

Square of their sum = ( x + x + 2 )²

Sum of their square = x² + ( x + 2 )²

According to the question -

→ ( x + x + 2 )² - [ x² + ( x + 2 )²] = 126

→ (2x + 2)² - x² - x² - 4 - 2 ( x ) ( 2 ) = 126

→ 4x² + 4 + 2 ( 2x )( 2 ) - 2x² - 4 - 4x = 126

→ 4x² + 4 + 8x - 2x² - 4 - 4x = 126

→ 4x² - 2x² + 8x - 4x + 4 - 4 = 126

→ 2x² + 4x = 126

→ 2x² + 4x - 126 = 0

→ 2 ( x² + 2x - 63 ) = 0

→ x² + 2x - 6x = 0

→ x² + 9x - 7x - 63x = 0

→ x ( x + 9 ) - 7 ( x + 9 ) = 0

→ ( x - 7 ) ( x + 9 ) = 0

x = 7 , - 9

As we need positive integer,

x = 7

Numbers = 7 and 9