The difference between the squares of two consecutive integers is 11. Find the integers?

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  • Como
    Lv 7
    3 weeks ago

    :-

    ( x + 1 )² - x² = 11

    x² + 2x + 1 - x² = 11

    2x = 10

    x = 5

    Integers are 5 and 6

  • 3 weeks ago

    The answer is 5 and 6 where the square of 5 = 25 and the square of 6 = 36, so the remainder would be 11.

  • 3 weeks ago

    The difference between the squares of two consecutive integers is 11.

    (x + 1)^2 - x^2 = 11

    (2x + 1) = 11

    x = 5

    The integers are 5 and 6.

  • Philip
    Lv 6
    3 weeks ago

    Let 2 consecutive integers be n, n+1. Then (n+1)^2 - n^2

    = 11, ie., 2n+1 = 11, ie., n = 5 & consecutive integers are

    5 and 6. 6^2 - 5^2 = 36 - 25 = 11.

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  • sepia
    Lv 7
    3 weeks ago

    The difference between the squares of two consecutive integers is 11.

    (a^2 - b^2) = (a + b)(a - b)

    ((n - 1) + n))((n - 1) - n)) = 11

    (2n - 1)(-1) = 11

    1 - 2n = 11

    2n = -10

    The two consecutive integers are -6 and -5.

  • 3 weeks ago

    The difference between n² and (n+1)² is |2n+1|.

    The solutions for |2n+1|=11 are 5 and -6.

    The consecutive integers are 5,6 and -6,-5.

  • 3 weeks ago

    I am not sure why the fox said this

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