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3 November, 14:37

Consider two consecutive positive integers such that the square of the second integer added to 3 times the first is equal to 37.

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  1. 3 November, 16:27
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    Am I supposed to consider or solve? Let's solve.

    Let's call our integers x and x+1

    square of the second integer added to 3 times the first is 37

    (x+1) ² + 3x = 37

    x² + 2x + 1 + 3x - 37 = 0

    x² + 5x - 36 = 0

    (x+9) (x-4) = 0

    Answer: x = - 9 or x=4

    Check:

    x=-9

    (-9+1) ² + 3 (-9) = 64 - 27 = 37 good

    x=4

    (4+1) ² + 3 (4) = 25 + 12 = 37 good

    Two solutions.
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