Question

Find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer 1

Let one of the natural numbers be ‘a’ 

Given, the numbers differ by 3. 

⇒ 2^nd number = a + 3 

⇒ a² + (a + 3)² = 117 

⇒ a² + a² + 6a + 9 = 117 

⇒ a² + 3a – 54 = 0 

⇒ a² + 9a – 6a – 54 = 0 

⇒ a(a + 9) – 6(a + 9) = 0 

⇒ (a – 6)(a + 9) = 0 

⇒ a = 6, - 9 

Thus, the numbers are 6, 9