SOLUTION: Find two consecutive odd integers such that their product is 35 more than 8 times their sum.

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Question 1014284: Find two consecutive odd integers such that their product is 35 more than 8 times their sum.
Found 2 solutions by KMST, MathTherapy:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
= the smaller of the two odd integers
= the next odd integer
= the product of those two consecutive odd integers
= the sum of those two consecutive odd integers\
We translate "their product is 35 more than 8 times their sum" as
.
That is the equation we have to solve, to find (and ).

If we find a solution that is even, or negative, or not an integer at all, we discard it.
If we find a solution that is an odd integer, that is the answer to the problem.
If we do not find a solution that is an odd integer, the problem has no solution.








That is a quadratic equation that can be solved 3 ways.
BY COMPLETING THE SQUARE:


--->--->--->
BY FACTORING:


--->--->
BY USING THE QUADRATIC FORMULA:


The quadratic formula, to find the solution to is
.
In the case of , , and
--->--->
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Find two consecutive odd integers such that their product is 35 more than 8 times their sum.
Let smaller integer be S

Then larger is: S + 2

We then get: S(S + 2) = 8(S + S + 2) + 35









(S - 17)(S + 3) = 0 ------- Factoring the above trinomial

Smaller integer, or S = 17          OR          S = - 3



If the smaller integer = , then larger integer = 17 + 2, or     

However, if the smaller integer = , then larger integer = - 3 + 2, or  

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