find three consecutive multiples of five such that the product of the first and second numbers minus seven times the 3rd one is 125?

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  • 1 month ago
    Best answer

    find three consecutive multiples of five such that the product of the first and second numbers minus seven times the 3rd one is 125?

    call the numbers x, x+5, x+10

    x(x+5) – 7(x+10) = 125

    x² + 5x – 7x – 70 – 125 = 0

    x² – 2x – 195 = 0

    quadratic equation:

    to solve ax² + bx + c = 0

    x = [–b ± √(b²–4ac)] / 2a

    x = [2 ± √(4+4•196)] / 2

    x = [2 ± √(784)] / 2

    x = [2 ± 28] / 2

    x = 1 ± 14

    x = 15, –13

    so answer is 15, 20, 25

    check

    15(20) – 7(25) = 125

    300 – 175 = 125

    ok

  • Como
    Lv 7
    1 month ago

    5x ( 5x. + 5 ) - 7 ( 5x + 10 ) = 125

    25x² + 25x - 35 x - 70 = 125

    25x² - 10 x - 195 = 0

    5x² - 2x - 39 = 0

    ( 5x + 13 ) ( x - 3 ) = 0

    x = 3 is acceptable integer

    Three consecutive integers are 15 , 20 and 25

    Check

    (15 x 20) - 175 = 300 - 175 = 125 ____as required

  • 1 month ago

    Three consecutive multiples of five such that

    the product of the first and second numbers

    minus seven times the 3rd one is 125:

    5(x - 1)5x - 7(5(x + 1)) = 125

    x = 4

    15, 20 and 25

  • rotchm
    Lv 7
    1 month ago

    let the three multiples be 5(n-1), 5n, 5(n+1).

    You Are told that 5(n-1) * 5n -7*5(n+1) = 125

    Solve for n.

    Done.

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