SOLUTION: The product of two positive consecutive integers is 41 more than their sum. Find the integers.

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Question 283042: The product of two positive consecutive integers is 41 more than their sum. Find the integers.
Found 2 solutions by unlockmath, MathTherapy:
Answer by unlockmath(1688) (Show Source):
You can put this solution on YOUR website!
Hello,
Let's have x be a positive number and the next to be x+2. Now we can set up an equation to be:
x(x+2)=x+(x+2)+41 Expand this out to:
x^2+2x=2x+43 Subtract 2x from both sides to get:
x^2=43 Square root both side to get:
x=+-sq rt 43
There we go.
Make sense?
RJ
www.math-unlock.com

Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
The product of two positive consecutive integers is 41 more than their sum. Find the integers.

Let the first number be F

Then the second = F + 1

Their product = F(F + 1) = , and their sum is: F + F + 1 = 2F + 1

Since their product is 41 more than their sum, then we have:

(F + 6)(F - 7) = 0

Ignore F + 6 = 0, because F = - 6, and we're looking for a positive number. Therefore, F, or the 1st number is 7, and so, the numbers are