SOLUTION: Find two consecutive odd integers such that their product is 143 more than 5 times their sum.

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Question 1163850: Find two consecutive odd integers such that their product is 143 more than 5 times their sum.
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The th odd integer is where is any integer. The next consecutive integer is 2 more:



Solve for , then calculate and

Hint: The equation is a quadratic. There are two solutions.

John

My calculator said it, I believe it, that settles it

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing it for you, I'll do one just like it:

Find two consecutive odd integers such that their product is 1 less than 6 times their sum.
Let the smaller one be x and larger be x+2.

x(x+2) = 6[x+(x+2)]-1

x²+2x = 6[x+x+2]-1
x²+2x = 6[2x+2]-1
x²+2x = 12x+12-1
x²+2x = 12x+11
x²-10x-11 = 0
(x-11)(x+1) = 0

x-11=0;  x+1=0
   x=11;   x=-1

One answer is 11 and 13, the other answers are -1 and +1

Checking 11 and 13.  Their product is 143, which is 1 less than
6 times their sum, which is 6 times 11+13=24 and 6×24 = 144, and
143 is 1 less than 144. So 11 and 13 checks.

Checking -1 and 1.  Their product is -1, which is 1 less than
6 times their sum, which is 6 times -1+1 = 0 and 6×0 = 0, and
-1 is 1 less than 0. So -1 and 1 checks. 

Now use this as a guide and solve yours.  This problem has a
"less than" and yours has a "more than".  So you'll have to add
instead of subtract.

Edwin