SOLUTION: there are three consecutive integers. the product of the first two plus the third integer is equal to 101

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Question 586422: there are three consecutive integers. the product of the first two plus the third integer is equal to 101
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the integers be n, n+1, and n+2.
The product of the first two is n%28n%2B1%29=n%5E2%2Bn
That plus the third integer is n%5E2%2Bn%2Bn%2B2=n%5E2%2B2n%2B2
If n%5E2%2B2n%2B2=101 --> n%5E2%2B2n=99 (subtracting 2 from both sides).
Now we can complete the square to solve the quadratic equation:
n%5E2%2B2n=99 --> n%5E2%2B2n%2B1=99%2B1 --> %28n%2B1%29%5E2=100 --> n%2B1=10
The integers are 9, 10, and 11.
9%2A10%2B11=90%2B11=101
NOTE: There are other ways to solve that quadratic equation (factoring, the quadratic formula), but completing the square looked easier.