SOLUTION: One positive integer is 3 less than a second positive integer. The sum of the squares of the two integers is 65. Find both positive integers.

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Question 289589: One positive integer is 3 less than a second positive integer. The sum of the squares of the two integers is 65. Find both positive integers.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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One positive integer is 3 less than a second positive integer.
x = an integer
(x+3) = 2nd integer
:
The sum of the squares of the two integers is 65. Find both positive integers.
x^2 + (x+3)^2 = 65
FOIL
x^2 + x^2 + 6x + 9 = 65
:
2x^2 + 6x + 9 - 65 = 0
:
2x^2 + 6x - 56 = 0
Factors to
(2x + 14)(x - 4) - 0
Positive solution
x = 4 is the 1st integer
and
4 + 3 = 7 is the 2nd integer
;
:
Check
4^2 + 7^2 = 65