SOLUTION: The product of two consecutive integers is 71 more than their sum. Find the integers.

Question 79539: The product of two consecutive integers is 71 more than their sum. Find the integers.
Found 2 solutions by Earlsdon, fetter6:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let the two consecutive integers be x and (x+1). From the description you can write:
Simplify and solve for x.
Subtract 2x from both sides.
Subtract 72 from both sides.
Solve this quadratic equation by factoring.
Apply the zero products principle.
or so...
or
So, there are really two answers to this problem and both would apply since the prolem did not limit the solution to positive integers.
1st. solution: x = 9 and x+1 = 10
2nd. solution: x = -8 and x+1 = -7
Check:
1st. solution:
9*10 = 90 and
9+10+71 = 90
2nd. solution:
-8*(-7) = 56 and
(-8)+(-7)+71 = -15+71 = 56
So you see, both solutions work and both sets of numbers are integers.
Answer by fetter6(6) (Show Source): You can put this solution on YOUR website!
Let n and n+1 be the two consecutive integers. We know that

Multiply out the left hand side and simplify the right hand side.
.
Now set the equation to 0.

Factor the left hand side.

So n=9,-8.
So we have two solutions:
1. n=9, n+1=10 solution pair (9,10)
2. n=-8, n+1=-7 solution pair (-8,-7)
Check your answer with both of these to convince yourself that you are right.
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