SOLUTION: The product of two consecutive odd integers is 6 more than 5 times the smaller number, find both numbers

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Question 1026395: The product of two consecutive odd integers is 6 more than 5 times the smaller number, find both numbers
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be n and n+2
n(n+2) = 5n+6

n%5E2%2B2n+=5n%2B6

n%5E2-3n-6=0

n%5E2-3n%2B2n-6=0

n(n-3)+2(n-3)=0
(n-3)(n+2)=0
n=3 or n=-2
odd number so n=3
3,5 are the numbers

Answer by ikleyn(53646) About Me  (Show Source):
You can put this solution on YOUR website!
.
The product of two consecutive odd integers is 6 more than 5 times the smaller number, find both numbers
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        The "solution" in the post by @mananth is INCORRECT,
        since he incorrectly solved the quadratic equation.

        See my reasoning below. 


Let 'n' be the smaller of the two odd consecutive integers.
Then the greater add consecutive integer is (n+2).


Our equation is

    n*(n+2) = 5n + 6,

    n^2 + 2n = 5n + 6,

    n^2 - 3n - 6 = 0.


The discriminant is  d =  b^2 - 4ac = (-3)^2 - 4*1*(-6) = 9 + 24 = 33.


sqrt%28d%29 = sqrt%2833%29  is not an integer or a rational number, 
so integer 'n' as described in the problem, does not exist.


ANSWER.  As posed/printed/presented in the post, the problem HAS NO solution.

Solved correctly.