SOLUTION: The product of the first and third of three consecutive integers is 3 more than 3 times the second integer. Find the integers.

Question 643354: The product of the first and third of three consecutive integers is 3 more than 3 times the second integer. Find the integers.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Let x = the first integer.
Then the second integer would be x+1
and the third integer would be x+2

Using these expressions for the three integers we can translate "The product of the first and third of three consecutive integers is 3 more than 3 times the second integer." into:

Now we solve this. Quadratic inequalities are solved in much the same way that quadratic equations. First we simplify each side:

and then get one side to be zero. Subtracting 3x and 3 from each side we get:

Next we factor (or use the Quadratic Formula). Note: This will tell us the x's that make the left side equal to zero. We are interested in the x's that make the left side greater than zero. The x's that make the left side zero will help us fine the x's that make it greater than zero.

The left side will not factor (at least not in a normal way). So we must use the Quadratic Formula:

Simplifying...

which is short for
or

Remember, these are the x's that make zero. (BTW: This means that will factor into These two x values are the only two x values that will make zero. All other x's result in positive or negative values for . These two x values divide the other x values into three groups:
1) Those larger than
2) Those less than
3) Those in between and

To find our solution, we need to figure out which of the above groups will make positive. We will pick a number from each group to see if it works. (If you have trouble figuring a number for a group, use your calculator to get a decimal value for the x's we found earlier.)

For group 1, 10 is clearly a number in that group:

So group 1 works, It makes greater than zero.

For group 2, -10 is clearly a number in that group:

So group 2 works, It makes greater than zero.

For group 3, 0 is clearly a number in that group:

So group 3 does not work. It makes less than zero.

So our solutions come from groups 1 and 2:
1) Those larger than
2) Those less than
Since we are only interested in integer solutions, we only want integers that are greater than or less then . (Again, use decimals if you can't figure this out on mentally.) So our solutions are:
all integers greater than or equal to 3 (3, 4, 5, ...); or
all integers less than or equal to -2 (-2, -3, -4, ...)
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