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-   -   The product of two consecutive even integers is 224. Find the integers (https://www.askmehelpdesk.com/showthread.php?t=208297)

  • Apr 21, 2008, 08:18 PM
    pearmansa473
    the product of two consecutive even integers is 224. Find the integers
    :confused: what r the integers
  • Apr 21, 2008, 09:18 PM
    KISS
    Not too bad. Any number multiplied by 2 will be even. 1*2, 2*2, 3*2 etc.
    Consecutive even integers differs by 2 and defined by a*2 and (a+1)*2

    So (a*2)*(a+1)*2 = 224 [fixed - courtesy of ISneezeFunny, thanks]

    I1= a*2
    I2= (a+1)*2

    Just a dumb check: suppose a =3; I1=6, I2=8. They are consecutive even integers.

    Solve for a and then I1 and I2

    I agree. A little difficult, but you can't have 1 equation and 2 unknowns.
  • Apr 21, 2008, 11:40 PM
    ISneezeFunny
    To go along what keepitsimplestupid said, although (a*2) + (a+1)*2 is wrong, as that's addition and the question asks for product.

    2a is the first integer
    2a + 2 is the next even integer.

    (2a)(2a + 2) = 224

    4a^2 + 4a = 224

    a^2 + a = 56

    a^2 + a - 56 = 0

    (a + 8)(a - 7) = 0

    a = -8 or 7

    try -8

    first even integer = 2a = -16, second even integer is -14

    (-16)(-14) = 224, check.

    try 7

    first even integer = 2a = 14, second even integer is 16

    (14)(16) = 224, check
  • Apr 22, 2008, 06:19 AM
    ebaines
    It's really much simpler than all that. Just take the square root of 224, and the answer will be the two even numbers that "bracket" the square root.
  • Apr 23, 2008, 01:15 PM
    eeseely
    Eaines approach is the easiest way to solve this problem.

    That's the way I solved it.
  • Apr 27, 2008, 10:20 AM
    confusedbuzz
    I think the best way to do this is that you assign the smaller even number as X and the greater one as X + 2 since they are consecutive .
    therefore:

    x(x+2) = 224
    and you can solve this quadratically and you'll get the answer simply
    x = 14 , x = 16
    so the two integers are 14 and 16

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