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Question 21298: how do solve three consecutive even integers?
the question: find three consecutive even integers such that the product of the first and third minus the second is one more than 10 times the third.
my answer:10x
then thats where i get stuck can you help please.
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Let x = first even integer
x+2 = second even integer
x+4 = third even integer
Translate the following sentence:
The product of the first and third, minus the second is one more than 10 times the third.
x(x+4) - (x+2) = 10(x+4) + 1
Quadratic equation, so set it equal to zero:
This does not factor, so there are no solutions to this equation that are integers. The answer is NO SOLUTION!
On second thought, look at this problem again. Obviously there was no solution, from the very beginning. If I had been really paying attention, I would have seen that there is NO SOLUTION.
Consider this: The product of the FIRST and THIRD EVEN numbers would be EVEN, minus the SECOND, would still be EVEN. Now, for the last part of the equation, the TEN times the THIRD number is an EVEN number, but ONE more than this would be an ODD number. An EVEN number can't possibly equal an ODD number.
I suspect someone is playing with us on this one, and they caught me too--at first!! But we got them now, right???
R^2 at SCC
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