Question 750884
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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