SOLUTION: The sum of the squares of two consecutive positive numbers is 41. What is the smaller number?

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Question 312531: The sum of the squares of two consecutive positive numbers is 41. What is the smaller number?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
LET X & X+1 BE THE 2 NUMBERS.
41=X^2+(X+1)^2
41=X^2+X^2+2X+1
2X^2+2X+1-41=0
2X^2+2X-40=0
2(X^2+X+420)=0
2(X+5)(X-4)=0
X-4=0
X=4 ANS FOR THE SMALLER NUMBER.
PROOF:
41=4^2+(4+1)^2
41=16+5^2
41=16+25
41=41