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InterviewSolution
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290 TIFind two consecutive. positive odd wholembers whose sum of squares is 290. |
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Answer» Let one of the odd positive integer be xthen the other odd positive integer is x+2their sum of squares = x² +(x+2)² = x² + x² + 4x +4 = 2x² + 4x+ 4Given that their sum of squares = 290⇒ 2x² +4x + 4= 290⇒2x² +4x = 290-4 = 286⇒ 2x²+4x -286 = 0⇒ 2(x² +2x - 143) = 0⇒ x² + 2x - 143 = 0⇒ x² + 13x - 11x -143 = 0⇒ x(x+13) - 11(x+13) = 0⇒ (x-11) = 0 , (x+13) =0Therfore , x = 11 or -13We always take positive value of xSo , x = 11 and (x+2) = 11 + 2 = 13Therefore , the odd positive integers are 11 and 13 . |
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