Find two consecutive multiples of 3 whose product is 270.

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KnowledgeBoat · Accepted Answer

Given, Let two consecutive multiples of 3 be 3n and 3n + 3. Given, The product is 270. ⇒ 3n(3n + 3) = 270 ⇒ n(3n + 3) = ⇒ 3n2 + 3n = 90 ⇒ 3n2 + 3n - 90 = 0 ⇒ 3(n2 + n - 30) = 0 ⇒ n2 + n - 30 = 0 ⇒ n2 + 6n - 5n - 30 = 0 ⇒ n(n + 6) - 5(n + 6) = 0 ⇒ (n - 5)(n + 6) = 0 ⇒ (n - 5) = 0 or (n + 6) = 0 [Using zero-product rule] ⇒ n = 5 or n = -6 Case 1: n = 5 First multiple is 3n = 3(5) = 15 Second multiple is 3n + 3 = 3(5) + 3 = 15 + 3 = 18 Case 2: n = -6 First multiple is 3n = 3(-6) = -18 Second multiple is 3n + 3 = 3(-6) + 3 = -18 + 3 = -15 Hence, two consecutive multiples of 3 are 15 and 18 or -18 and -15.

Mathematics

Find two consecutive multiples of 3 whose product is 270.

Quadratic Equations

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