SOLUTION: Find two positive consecutive even integers whose product is 8 more than 5 times the larger integer.

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Question 720017: Find two positive consecutive even integers whose product is 8 more than 5 times the larger integer.
Answer by graphmatics(170) About Me  (Show Source):
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Let 2*n be the first even integer. Then the next even integer is 2*n+2. 8 more than 5 times the larger integer is 5*(2*n+2)+8. As this is the product of the consecutive integers we have that
(2*n)*(2*n+2)=5*(2*n+2)+8
2^2*n^2+4*n=10*n+10+8
2^2*n^2-6*n-18=0
2*n^2-3*n-9=0
Let's try to factor this expression
(2*n+3)*(n-3)=0
we get n=-3/2 and n=3. We must want the integer 3 so our even integers must be 6 and 8. Check that they work in the first equation.