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) About Me  (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