SOLUTION: Six less than the product of two consecutive positive integers is equal to six times their sum. Find the two integers.

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Question 953313: Six less than the product of two consecutive positive integers is equal to six times their sum. Find the two integers.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
n=first number; n+1=second number
((n)(n+1))-6=6(n+n+1)
n^2+n-6=6(2n+1)
n^2+n-6=12n+6 Subtract (12n+6) from each side.
n^2-11n-12=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 1n%5E2%2B-11n%2B-12+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-11%29%5E2-4%2A1%2A-12=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--11%2B-sqrt%28+169+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%28-11%29%2Bsqrt%28+169+%29%29%2F2%5C1+=+12
n%5B2%5D+=+%28-%28-11%29-sqrt%28+169+%29%29%2F2%5C1+=+-1

Quadratic expression 1n%5E2%2B-11n%2B-12 can be factored:
1n%5E2%2B-11n%2B-12+=+1%28n-12%29%2A%28n--1%29
Again, the answer is: 12, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-11%2Ax%2B-12+%29

ANSWER 1: The first positive integer is 12.
x+1=12+1=13 ANSWER 2: The second positive integer is 13.
CHECK:
(12*13)-6=6(12+13)
156-6=6(25)
150=150