#### Question

Find the consecutive numbers whose squares have the sum 85.

#### Solution

Let the two consecutive natural numbers be ‘x’ and ‘x + 1’

⇒ Given that the sum of their squares is 85.

Then by hypothesis, we get

𝑥^{2} + (𝑥 + 1)^{2} = 85

⇒ 𝑥^{2} + 𝑥^{2} + 2𝑥 + 1 = 85

⇒ 2𝑥^{2} + 2𝑥 + 1 − 85 = 0

⇒ 2𝑥^{2} + 2𝑥 + 84 = 0 ⇒ 2 [𝑥^{2} + 𝑥 − 42 ] = 0

⇒ 𝑥^{2} + 7𝑥 - 6𝑥 - 42 = 0 [by the method of factorisation]

⇒ 𝑥(𝑥 + 7) - 6(𝑥 + 7) = 0

⇒ (𝑥 - 6)(𝑥 + 7) = 0

⇒ 𝑥 = 6 or 𝑥 = 7

Case i: if x = 6 x + 1 = 6 + 1 = 7

Case ii: If x = 7 x + 1 = -7 + 1 = -6

∴ The consecutive numbers that the sum of this squares be 85 are 6,7 and -6, -7.

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#### APPEARS IN

Solution Find the Consecutive Numbers Whose Squares Have the Sum 85. Concept: Solutions of Quadratic Equations by Factorization.