#### Question

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

#### Solution

Let the numbers be x and x - 3

By the given hypothesis,

𝑥^{2} + (𝑥 - 3)^{2} = 117

⇒ 𝑥^{2} + 𝑥^{2} + 9 - 6𝑥 - 117 = 0

⇒ 2𝑥^{2} - 6𝑥 - 108 = 0

⇒ 𝑥^{2} - 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.

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#### APPEARS IN

Solution Let Us Find Two Natural Numbers Which Differ by 3 and Whose Squares Have the Sum 117. Concept: Solutions of Quadratic Equations by Factorization.