SOLUTION: find three consecutive integers such that the product of the two smaller inters is 2 more than ten times the largest integers.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find three consecutive integers such that the product of the two smaller inters is 2 more than ten times the largest integers.      Log On


   

Question 177196: find three consecutive integers such that the product of the two smaller inters is 2 more than ten times the largest integers.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the three consecutive integers be: x, (x+1) and (x+2)
x%2A%28x%2B1%29+=+10%28x%2B2%29%2B2 "The product of the two smaller integers (x and (x+1)) is 2 more than 10 times the larger integer (x+3).
x%5E2%2Bx+=+10x%2B20%2B2 Subtract 10x from both sides.
x%5E2-9x+=+22 Subtract 32 from both sides.
x%5E2-9x-22+=+0 Factor this quadratic equation.
%28x-11%29%28x%2B2%29+=+0 Apply the zero product rule.
x-11+=+0 or x%2B2+=+0 so...
x+=+11 or x+=+-2
Solutions:
11, 12, and 13
-2, -1, and 0
As you can see, there are two solutions, so let;'s check them both!
x = 11
%2811%29%2A%2812%29+=+10%2813%29%2B2
132+=+130%2B2
132+=+132 OK!
x = -2
%28-2%29%2A%28-1%29+=+10%280%29%2B2
2+=+2 OK!