SOLUTION: THE SUM OF THE SQUARES OF TWO CONSECUTIVE POSITIVE INTEGERS IS 41
41=n^2(n+1)^2
n^2+n^2+2n-40=0
n^2+2n-40=0
factor out 2
n^2+n-20=0
I tried different factors of -20 (n+1)(n-2
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: THE SUM OF THE SQUARES OF TWO CONSECUTIVE POSITIVE INTEGERS IS 41
41=n^2(n+1)^2
n^2+n^2+2n-40=0
n^2+2n-40=0
factor out 2
n^2+n-20=0
I tried different factors of -20 (n+1)(n-2
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Question 394704: THE SUM OF THE SQUARES OF TWO CONSECUTIVE POSITIVE INTEGERS IS 41
41=n^2(n+1)^2
n^2+n^2+2n-40=0
n^2+2n-40=0
factor out 2
n^2+n-20=0
I tried different factors of -20 (n+1)(n-20)=-19; (n+2)(n-10)=-8
(n+10)(n-2)=8 (n+20)(n-1)=19 i know the answers are supposed to be 2 and 4 so I know I am wrong but where? Answer by richard1234(7193) (Show Source):
