A natural number is greater than the other by 5. The sum of their squares is 73. Find those numbers.

10. A natural number is greater than the other  by 5. The sum of their squares is 73. Find those numbers.


Sol.  Let the second natural number be x.

∴  The first natural number is x + 5.

As per the given condition,

x2 + (x + 5)2 = 73

∴  x2 + x2 + 10x + 25 – 73 = 0

∴  2x2 + 10x – 48 = 0

Dividing by 2, we get

x2 + 5x – 24 = 0

∴  x2 – 3x + 8x – 24 = 0

∴ x (x – 3) + 8 (x – 3) = 0

∴ (x – 3) (x + 8) = 0

∴ x – 3 = 0 or x + 8 = 0

∴x = 3 or x = - 8

∵ the natural number cannot be negative

∴ x = 3

and x + 5 = 3 + 5 = 8


∴ the first natural number is 8 and the second is 3.