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Let one positive integer be x, so the other is x+3. The sum of the squares of the two integers is 89, so we have:

x2 + (x+3)2 = 89

x2 + (x2+6x+9) = 89

2x2 + 6x + 9 = 89

2x2 + 6x +9 - 89 = 0

2x2 + 6x - 80 = 0

x2 + 3x - 40 = 0

(X+8)(x-5)=0

x = 5 (the only positive integer)

The 2 positive integers are 5 & 8. Squaring both would be 25 & 64, which the sum is 89.

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Q: How can you solve the problem one positive integer is 3 greater than anotherthe sum of the squares of the two integers is 89find the integers?
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