SOLUTION: Find two consecutive integers whose product is 6 less than the square of the larger number.

Question 267374: Find two consecutive integers whose product is 6 less than the square of the larger number.
Answer by Ruli(21) (Show Source): You can put this solution on YOUR website!
The product of two consecutive integers = (x)(x+1)
6 less than the square of the larger number = (x+1)^ - 6
(x)(x+1) = (x+1)^ - 6
x^ + 1x = (x^ + 2x + 1) - 6
x^ + 1x = x^ + 2x - 5
x^ + 1x - x^ = x^ + 2x - 5 - x^
1x - 2x = 2x - 5 - 2x
-1x = -5
Divide each side by -1
x = 5
check:
(5)(5+1) = (5+1)^-6
(5)(6) = (6)^-6
30 = 36-6
30 = 30
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