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5 April, 07:38

Find three consecutive positive integers such that the product of the first and second is 37 less thanthe square of the third

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Answers (2)
  1. 5 April, 09:35
    0
    Consecutive integers are 1 apart

    they are x, x+1, and x+2

    produce of first and 2nd (x and x+1) is 37 less than square of third (x+2)

    (x) (x+1) = - 37 + (x+2) ²

    expand

    x²+x=x²+4x+4-37

    x²+x=x²+4x-33

    minu x² from both sides

    x=4x-33

    minus x both sides

    0=3x-33

    add 33 both sides

    33=3x

    divide by 3 both sides

    11=x

    x+1=12

    x+2=13

    the numbers are 11,12,13
  2. 5 April, 11:02
    0
    See the pic for the answer.
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