SOLUTION: The sum of two consectutive positive integers is 11 less than their product. find the integers.

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Question 37226: The sum of two consectutive positive integers is 11 less than their product.
find the integers.
Found 2 solutions by Paul, askmemath:
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the two consecutive integers be x and x+1
(x)+(x+1)=(x)(x+1)-11
2x=x^2+x-12
x^2-x-12=0
Solve that and you get:
x=4
4+1=5
Hence, the two numbers are 4 and 5.
Paul.

Answer by askmemath(368) About Me  (Show Source):
You can put this solution on YOUR website!
2 Consecuitve numbers = X and X+1 ( e.g. 3 and 4 are consecuitve and they differ by 1)
Sum = X +X+1 = 2X+1
Product = X(X+1) = X%5E2+%2BX
Sum less than product by 11 which means Sum+11 = Product
2x%2B1+%2B11+=+X%5E2+%2BX
Re-arranging
X%5E2+-X-12
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-12+=+0) has the following solutons:

x%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=%28-1%29%5E2-4%2A1%2A-12=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+4
x%5B2%5D+=+%28-%28-1%29-sqrt%28+49+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-1x%2B-12 can be factored:
1x%5E2%2B-1x%2B-12+=+1%28x-4%29%2A%28x--3%29
Again, the answer is: 4, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-12+%29