Question

Let us find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer 1

Let the numbers be x and x - 3

By the given hypothesis,

𝑥² + (𝑥 - 3)² = 117

⇒ 𝑥² + 𝑥² + 9 - 6𝑥 - 117 = 0

⇒ 2𝑥² - 6𝑥 - 108 = 0

⇒ 𝑥² - 3𝑥 - 54 = 0

⇒ 𝑥(𝑥 - 9) + 6(𝑥 - 9) = 0

⇒ (x - 9) (x + 6) = 0

⇒ x = 9 or x = -6

Considering positive value of x

x = 9, x - 3 = 9 - 3 = 6

∴ The two numbers be 9 and 6.