SOLUTION: What is the smallest of three consecutive positive numbers if the product of the first and third is one less than eight times the second?

Question 973229: What is the smallest of three consecutive positive numbers if the product of the first and third is one less than eight times the second?
Answer by crystal5213(2) (Show Source): You can put this solution on YOUR website!
The smallest number is equal to 7.
Explanation:
For this problem, I'll be using variables to represent the numbers: x, y, and z. But since they are consecutive, y=x+1, and z=x+2, so the numbers can be written as x, x+1, and x+2.
According to the problem, the product of the first and third number ( or ) is equal to one less than eight times the second number ( or )
We can set the problem up like this:

Now we can solve it like an algebraic equation instead of a word problem. Start by moving the variable to one side of the equation by subtracting 8x from both sides, resulting in:

Since there is a variable squared, we should use the quadratic formula. So we will have to subtract 7 as well, so that 0 will be on one side of the equal sign.

According to the quadratic equation, if then
Now, we plug in the numbers.

in this case, a equals one because there is no number before , so the coefficient must be 1. b is -6 since that is the coefficient for x, and c is -7.
Finally, after using a calculator to solve the equation, we get

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