SOLUTION: find two consecutive positive integers such that the square of the smaller is one more than eight times the larger.

Question 519917: find two consecutive positive integers such that the square of the smaller is one more than eight times the larger.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Two consecutive positive integers are defined by:
x
x+1
.
x^2 = 8(x+1) +1
.
x^2 = 8x +8 +1
.
x^2 -8x -9 = 0
.
factor
.
(x +1)(x -9) = 0
.
x = -1 or 9
.
Since we are limited to positive integers, which is 9.
.
Answer: The two consecutive positive integers are 9 and 10.
.
Check the answer.
.
9^2 = 81
8*10+1 = 81
correct
.
Done.
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