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 

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
 41/2 = 20.5

 apx

--> -4 & -5

 

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