Question

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

Answer 1

Let the numbers be x and x-3. Then,

x² + (x-3)2 = 117, 

⇒  x² + x² + 9 - 6x =117 

⇒ 2 x² -6x - 108 = 0 

⇒ x² - 3x - 54 = 0 

⇒ x² - 9x + 6x - 54 = 0 

⇒ x (x - 9) + 6(x - 9) = 0 

⇒ (x-9) (x+6 ) = 0 

⇒ x = 9  (As the number have to be natural number) 

Hence the numbers are 6 and 9. 

Answer 2

Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117
⇒ x² + x² + 6x + 9 = 117
⇒ 2x² + 6x + 9 – 1117 = 0
⇒ 2x² + 6x – 108 = 0
⇒ x² + 3x – 54 = 0   ...(Dividing by 2)
⇒ x² + 9x – 6x – 54 = 0
⇒ x(x + 9) –6(x + 9)  0
⇒ (x + 9)(x – 6) = 0
Either x + 9 = 0,
then x = –9,
but it is not a natural number.
or
x - 6 = 0,
then x = 6
∴ First natural number = 6
and second numberr = 6 + 3 = 9.