One positive integer is 3 3 less than twice another. The sum of their squares is 170 170 . Find the integers.?

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  • sepia
    Lv 7
    4 months ago

    One positive integer is 3 less than twice another.

    The sum of their squares is 170 .

    The integers are 11 and 7.

  • david
    Lv 7
    4 months ago

    n = one integer

    2n - 33 = 2nd integer

    =========================

    n^2 + (2n - 33)^2 = 170,170

    n^2 + 4n^2 - 264n + 1089 = 170,170

    5n^2 - 264n - 169081 = 0

    n^2 - 52.8n = 33816.2 .... complete the square (52.8 / 2)^2 = 696.96

    n^2 - 52.8n + 696.96 = 33816.2 + 696.96 = 34513.16

    (n - 26.4)^2 = +/- 185.7771784

    n = 212.1771784 <<< not an integer,

    something is wrong with this question ... As it is worded it is impossible. I believe you nade ab error entering the question. Try again.

  • 4 months ago

    why are you repeating numbers ??

    assuming this is "One positive integer is 3 less than twice another. The sum of their squares is 170"

    x = 2y – 3

    x² + y² = 170

    (2y – 3)² + y² = 170

    4y² + 9 – 12y + y² = 170

    5y² – 12y – 161 = 0

    solving, we get y = 7 or –23/5

    skipping the second result

    y = 7

    x = 2y – 3 = 2•7 – 3 = 11

    check

    7² + 11² = 170

    49 + 121 = 170 ok

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