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) About Me  (Show Source):
You can put this solution on YOUR website!
n= the smaller of the two odd integers
n%2B2= the next odd integer
n%28n%2B2%29= the product of those two consecutive odd integers
n%2B%28n%2B2%29= the sum of those two consecutive odd integers\
We translate "their product is 35 more than 8 times their sum" as
n%28n%2B2%29=8%28n%2B%28n%2B2%29%29%2B35 .
That is the equation we have to solve, to find n (and n%2B2 ).

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.

n%28n%2B2%29=8%28n%2B%28n%2B2%29%29%2B35
n%5E2%2B2n=8%28n%2Bn%2B2%29%2B35
n%5E2%2B2n=8%28n%2Bn%2B2%29%2B35
n%5E2%2B2n=8%282n%2B2%29%2B35
n%5E2%2B2n=16n%2B16%2B35
n%5E2%2B2n-16n=16%2B35
n%5E2-14n=51
That is a quadratic equation that can be solved 3 ways.
BY COMPLETING THE SQUARE:
n%5E2-14n=51
n%5E2-14n%2B49=51%2B49
%28n-7%29%5E2=100--->system%28n-7=10%2C%22or%22%2Cn-7=-10%29--->system%28n=10%2B7=17%2C%22or%22%2Cn=-10%2B7=-3%29--->highlight%28system%28n=17%2Cn%2B2=19%29%29
BY FACTORING:
n%5E2-14n=51
n%5E2-14n-51=0
%28n%2B3%29%28n-17%29=0--->system%28n=17%2C%22or%22%2Cn=-3%29--->highlight%28system%28n=17%2Cn%2B2=19%29%29
BY USING THE QUADRATIC FORMULA:
n%5E2-14n=51
n%5E2-14n-51=0
The quadratic formula, to find the solution to ax%5E2%2Bbx%2Bc=0 is
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ .
In the case of n%5E2-14n-51=0 , system%28a=1%2Cb=-14%2Cc=-51%29 , and
--->system%28x=%2814%2B20%29%2F2=34%2F2=17%2C%22or%22%2Cx=%2814-20%29%2F2=%28-6%29%2F2=-3%29--->highlight%28system%28n=17%2Cn%2B2=19%29%29

Answer by MathTherapy(10552) About Me  (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%5E2+%2B+2S+=+8%282S+%2B+2%29+%2B+35

S%5E2+%2B+2S+=+16S+%2B+16+%2B+35

S%5E2+%2B+2S+-+16S+-+51+=+0

S%5E2++-+14S+-+51+=+0

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

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



If the smaller integer = highlight%28highlight_green%28highlight%2817%29%29%29, then larger integer = 17 + 2, or highlight%28highlight_green%28highlight%2819%29%29%29    

However, if the smaller integer = highlight%28highlight_green%28highlight%28-+3%29%29%29, then larger integer = - 3 + 2, or highlight%28highlight_green%28highlight%28-+1%29%29%29