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Find two consecutive odd positive integers, sum of whose squares is 970. |
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Answer» Let the consecutive odd positive integers be ‘a’ and a + 2 ⇒ a2 + (a + 2)2 = 970 ⇒ 2a2 + 4a – 966 = 0 ⇒ a2 + 2a – 483 = 0 ⇒ a2 + 23a – 21a – 483 = 0 ⇒ a(a + 23) – 21(a + 23) = 0 ⇒ (a – 21)(a + 23) = 0 Thus, a = 21 Consecutive odd positive integers are 21, 23 |
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