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Find two consecutive numbers whose squares have the sum of 85 |
Answer» Let the consecutive numbers be ‘a’ and a + 1. Given, sum of squares is 85 ⇒ a2 + (a+ 1)2 = 85 ⇒ a2 + a2 + 2a + 1 = 85 ⇒ a2 + a – 42 = 0 ⇒ a2 + 7a – 6a – 42 = 0 ⇒ a(a + 7) – 6(a + 7) = 0 ⇒ (a – 6)(a + 7) = 0 ⇒ a = 6, -7 Numbers are, 6, 7 or -7, -6 |
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