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Question 1163695: The sum of the square of 2 consecutive natural numbers is 41. Find the numbers
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the two consecutive numbers are x and x + 1
therefore:
x^2 + (x + 1)^2 = 41
simplify to get:
x^2 + x^2 + 2x + 1 = 41
subtract 41 from both sides and combine like terms to get:
2x^2 + 2x - 40 = 0
divide both sides of this equation by 2 to get:
x^2 + x - 20 = 0
factor this quadratic equation to get:
(x - 4) * (x + 5) = 0
solve for x to get:
x = 4 or x = -5
since x is a natural number, than the only possible solution is x = 4
when x = 4, x + 1 = 5
your two natural numbers are 4 and 5.
x^2 + (x + 1)^2 = 41 becomes:
4^2 + 5^2 = 41 which becomes:
16 + 25 = 41 which becomes:
41 = 41, confirming the solution is correct.
your solution is that the numbers are 4 and 5.
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