SOLUTION: the sum of the squares of two consecutive positive integers is nine more than 8 times the smaller. Find the integers.

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Question 470907: the sum of the squares of two consecutive positive integers is nine more than 8 times the smaller. Find the integers.
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
Let x and (x+1) represent the two consecutive positive integers
Question states***
x^2 + (x+1)^2 = 8x + 9
x^2 + x^2 + 2x + 1 = 8x +9
2x^2 -6x - 8 = 0
x^2 - 3x - 4 = 0
factoring
(x-4)(x+1) = 0
(x-4) = 0 x = 4 Integers are 4,5
(x+1) = 0 x = -1 Toss out, looking for positive pair
CHECKING our Answer***
16 + 25 = 41 = 32+9

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