Question 1057459: Find three consecutive negative integers such that the product of the first two is 26 greater than the square of the third.
For some reason I end up with a positive integer after working it out, any help?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find three consecutive negative integers such that the product of the first two is 26 greater than the square of the third.
let our numbers be:(n-2), (n-1), n
(n-2)(n-1) = n^2 + 26
FOIL the left
n^2 - 3n + 2 = n^2 + 26
combine like terms
n^2 - n^2 - 3n = 26 - 2
-3n = 24
n = 24/-3
n = -8
:
the three number -10, -9, -8
:
:
see if the works
-10 *-9 = -8^2 + 26
+90 = +64 + 26
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