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Question 814774: find three consecutive odd positive integers such that 2 times the sum of all three is 7 more than the product of the first and second integers
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! j = i + 2
k = i + 4
2(i + j + k) = ij + 7
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2i + 2(i + 2) + 2(i + 4) = i(i + 2) + 7
2i + 2i + 4 + 2i + 8 = ii + 2i + 7
ii - 4i - 5 = 0
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the above quadratic equation is in standard form, with a=1, b=-4, and c=-5
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to solve the quadratic equation, plug this:
1 -4 -5
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the roots of the quadratic are:
5
-1
the problem rules outlaw the negative root, so:
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Answer:
i = 5
j = 7
k = 9
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