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

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) (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) (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) =
Sum less than product by 11 which means Sum+11 = Product

Re-arranging

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=49 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4, -3. Here's your graph:

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