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6 October, 10:41

Three consecutive odd integers are such that the square of the third integer is 9 less than the sum of the squares of the first two. One solution is negative - 3 , - 1 , and 1. Find three other consecutive odd integers that also satisfy the given conditions

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  1. 6 October, 14:33
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    Hello,

    Let's assume a the first odd integer

    a+2 the second,

    a+4 the third.

    (a+4) ²=a² + (a+2) ²-9

    ==> a²+8a+16=a²+a²+4a+4-9

    ==> a²-4a-21=0

    Δ=16+4*21=10²

    ==> a = (4-10) / 2=-3 or a = (4+10) / 2=7

    sol n°1 = (-3; -1; 1)

    sol n°2 = (7; 9; 11)
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