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

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

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

Answer»

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.