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anonymous

  • 5 years ago

Find four consecutive intergers such that the product of the 1st and 3rd is 25 greater than of -13 and the fourth

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  1. mathteacher1729
    • 5 years ago
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    Weinburgerd, what part of this problem is giving you difficulty? I can help in a bunch of ways, but It helps if I know what's giving you trouble. :)

  2. anonymous
    • 5 years ago
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    Ok Its been a while since i have taken math and I was asked to help. This is what we came up with so far.

  3. anonymous
    • 5 years ago
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    n n-1 n-2 n-3

  4. anonymous
    • 5 years ago
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    n(n-2)=25-13(n-3) n^2 -2n= 25-13n +39 n^2 - 2n= 64 -13n n^2= 64-11n how do we find what n equals?

  5. anonymous
    • 5 years ago
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    I set the four consecutive integers to n, n+1,n+2 & n+3 n(n+2)=25-13(n+3) n^2 -2n= 25-13n -39 n^2 - 2n= -13n-14 n^2= -15n-14 n^2+15n+14=0 use the quadratic equation to solve n=-14 or -1

  6. anonymous
    • 5 years ago
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    Oops cut and paste and forgot to change the sign n^2 + 2n= 25-13n -39 n^2 + 2n= -13n-14 n^2= -15n-14 n^2+15n+14=0 use the quadratic equation to solve n=-14 or -1

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