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Solution of this questionFind two consecutive positive integers sum of whose square is 365. |
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Answer» 14 Let be the two consecutive +ve integers are x and ( x+1)According to the question — x² + (x+1)² = 365 x² + x² + 2x + 1 = 365 [using (a+b)² = a² +2ab+ b²]2x² +2x = 365 - 12x² + 2x - 364 = 0Dividing above equation by 2 ,we get —x² + x - 182 = 0 x² + (14-13)x - 182 = 0x² + 14x - 13x - 182 = 0x(x+14) -13(x+14) = 0(x + 14)(x - 13) = 0So , (x = -14) & (x = 13) Since , positive cosecutive integer can never be negative .Hence, two consecutive positive integers are : x = 13 & (x + 1) = 14 Ans. Ans. |
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