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2 November, 02:24

Find three consecutive integers such that the product of the two largest is 20 more than the square of the smallest number

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Answers (1)
  1. 2 November, 04:09
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    Givens

    the smallest integer is x

    The second smallest integer is x + 1

    The largest integer is x + 2

    Equations

    (x + 1) (x + 2) = x² + 20

    Solution

    x² + x + 2x + 2 = x² + 20 Subtract x² from both sides

    3x + 2 = 20 Subtract 2 from both sides.

    3x = 20 - 2

    3x = 18 Divide by 3

    x = 18/3

    x = 6

    The three numbers are 6,7 and 8

    Check

    7*8 = 6^2 + 20

    56 = 36 + 20

    56 = 56

    The equation's solution does check.
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