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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