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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
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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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Weinburgerd, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0n(n2)=2513(n3) n^2 2n= 2513n +39 n^2  2n= 64 13n n^2= 6411n how do we find what n equals?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I set the four consecutive integers to n, n+1,n+2 & n+3 n(n+2)=2513(n+3) n^2 2n= 2513n 39 n^2  2n= 13n14 n^2= 15n14 n^2+15n+14=0 use the quadratic equation to solve n=14 or 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oops cut and paste and forgot to change the sign n^2 + 2n= 2513n 39 n^2 + 2n= 13n14 n^2= 15n14 n^2+15n+14=0 use the quadratic equation to solve n=14 or 1
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