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

Question 750884: The sum of the squares of two consecutive positive numbers is 41. What is the smaller number?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
x^2 + (x+1)^2 = 41

x^2 + x^2 + 2x+1 = 41

2x^2 + 2x+1 = 41

2x^2 + 2x+1 - 41 = 0

2x^2 + 2x - 40 = 0

2(x^2 + x - 20) = 0

2(x + 5)(x - 4) = 0

x + 5 = 0 or x - 4 = 0

x = -5 or x = 4

Ignore -5 (since the two numbers are both positive). So one number is 4 and the other number is 5.

Answer: The smaller number is 4.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of the squares of two consecutive positive numbers is 41. What is the smaller number?
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1st: x
2nd: x+1
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Equation:
x^2 + x^2+2x+1 = 41
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2x^2 + 2x - 40 = 0
x^2 + x -20 = 0
(x+5)(x-4) = 0
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Positive solution:
x = 4
x+1 = 5
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Cheers,
Stan H.
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