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- anonymous

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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- anonymous

- chestercat

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- mathteacher1729

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. :)

- anonymous

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.

- anonymous

n
n-1
n-2
n-3

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- anonymous

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?

- anonymous

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

- anonymous

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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