SOLUTION: find three consecutive intergers such that the product of the second and third is 56

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find three consecutive intergers such that the product of the second and third is 56      Log On


   

Question 560148: find three consecutive intergers such that the product of the second and third is 56
Found 2 solutions by josmiceli, Alan3354:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the integers n, n%2B1, and n%2B2
given:
+%28+n%2B1+%29%2A%28+n%2B2+%29+=+56+
+n%5E2+%2B+3n+%2B+2+=+56+
+n%5E2+%2B+3n+-+54+=+0+
Use quadratic formula
n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+3+
+c+=+-54+
n+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A1%2A%28-54%29+%29%29%2F%282%2A1%29+
n+=+%28-3+%2B-+sqrt%28+9+%2B+216+%29%29+%2F+2+
n+=+%28-3+%2B-+sqrt%28+225+%29%29+%2F+2+
n+=+%28+-3+%2B+15+%29+%2F+2+
n+=+6+ ( I'll ignore the negative root of 225 )
+n+%2B+1+=+7+
+n+%2B+2+=+8+
The consecutive numbers are 6,7,and 8

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive intergers such that the product of the second and third is 56
---------------
sqrt(56) =~ 7.5, the middle
--> 7*8 = 56
--> 6, 7 & 8
-9, -8 & -7