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    johanna35's Avatar
    johanna35 Posts: 2, Reputation: 1
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    #1

    Apr 22, 2012, 01:35 PM
    quadratic word problem help
    find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7 times the second. Please can some one solve this problem so I can see how it's done?
    johanna35's Avatar
    johanna35 Posts: 2, Reputation: 1
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    #2

    Apr 22, 2012, 03:49 PM
    How come no one has aswered my question
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    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Apr 23, 2012, 06:13 AM
    The three consecutive odd integers are n, n+2 and n+4. The problem states that "the product of the first and third" - which is n(n+4) - is equal to "4 less than 7 times the second" - which is 7(n+2)-4. So you have:

    n(n+4) = 7(n+2)-4

    Can you take it from here? You will get two possible answers for n, but only one of which is odd.
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