SOLUTION: find three consecutive integers such that 3 times the sqaure of the first exceeds the product of the other two by 25

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Question 127004: find three consecutive integers such that 3 times the sqaure of the first exceeds the product of the other two by 25
Answer by solver91311(24713) About Me  (Show Source):
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First integer: x
Second integer: x + 1
Third integer: x + 2

3 times the square of the first: 3x%5E2
The product of the other two: %28x%2B1%29%28x%2B2%29=x%5E2%2B3x%2B2

3 times the square of the first exceeds the product of the other two by 25:3x%5E2=%28x%5E2%2B3x%2B2%29%2B25

3x%5E2-x%5E2-3x-27=0

2x%5E2-3x-27

%282x-9%29%28x%2B3%29=0

So x=9%2F2 or x=-3. We can exclude the first root because it is not an integer. So the first integer is -3, the second would be -2, and the third, -1.

Check:
3%28-3%29%5E2=27
%28-1%29%28-2%29%2B25=27, answer checks