SOLUTION: how would I find the product of two consecutive integers with the sum being 71 more than the integers?

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Question 268563: how would I find the product of two consecutive integers with the sum being 71 more than the integers?
Found 2 solutions by solver91311, unlockmath:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Your question doesn't make sense as posed. What does "71 more than the integers" mean?

If your problem were:

Find the product of two consecutive integers with the sum being 71.

I can do that.

Let represent the smaller integer. Then the next consecutive integer must be

So, if the sum is 71:



Solve for :







So the integers are 35 and 36. Multiply 35 times 36 to get the product.

WRONG! Got an update on the actual problem. It reads:

The product of two consecutive integers is 71 more than their sum. That means that unlockmath (see below) has the correct solution.

John

Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
Let x be one integer and x+1 be the next. Now we can set up the following equation: (Note: I'm assuming your problem is "more the the integers added" otherwise it doesn't make sense)
x(x+1)=x+(x+1)+71 Rewritten as:
x^2+x=2x+72 Subtract 2x and 72 from both sides to get:
x^2-x-72=0 Now we can factor this to be:
(x-9)(x+8)=0 Solve for x:
x=9
x=-8
There we go, the integers would be 9 and 10 or -8 and -7.
Make sense?
RJ
Check out a book I wrote at:
www.math-unlock.com