Let the numbers be x and x – 3, as its given the number differ by 3.
From the question, it’s given that sum of squares of these numbers is 117.
x2 + (x – 3)2 = 117
⇒ x2 + x2 + 9 – 6x – 117 = 0
⇒ 2x2 – 6x – 108 = 0
⇒ x2 – 3x – 54 = 0
Solving for x by factorization method, we have
⇒ x2 – 9x + 6x – 54 = 0
⇒ x(x – 9) + 6(x – 9) = 0
⇒ (x – 9)(x + 6) = 0
Now, either x – 9 = 0 ⇒ x = 9
Or, x + 6 = 0 ⇒ x = – 6
Considering only the positive value of x as natural numbers are always positive i.e, x = 9.
So, x – 3 = 6.
Thus, the two numbers are 6 and 9 respectively.
