Question 412046
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Hi
two consecutive negative integers
Let x and (x+1) represent the two consecutive 'negative' integers
Question states***
x^2 + (x+1)^2 = 41
Solving for x
 2x^2 + 2x + 1 = 41
 2x^2 + 2x - 40 = 0
  x^2 + x - 20 = 0
factoring
 (x-4)(x+5)= 0 Note:SUM of the inner product(-4x) and the outer product(5x) = x
 (x-4)=0  x = 4: Extraneous solution, question states 'negative' integers
 (x+5)= 0 x = -5  The two consecutive 'negative' integers are -5,-4

CHECKING our Answer***
 25 + 16 = 41