Algebra II : Median
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Example Question #81 : Median
There are three numbers. Say that one of the numbers represented is 
Let's interpret the problem. One number is 
Remember when foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables.
Remember, we need to find two terms that are factors of the c term that add up to the b term. We have:
Solve for 
We are not done as the problem asks for median of the set. If we plug in 
Example Question #5 : Median
There is a table of flowers prepared for sale. Twelve flowers are 
The easiest way to do this is first to find the total number of flowers:
Now, the median element is the "middle" term. To find the middle, you can divide 21 by 2:
Since you have an odd number of elements, you unsurprisingly get a fraction. This means that there 10 items to the left of the median and 10 to its right. The 11th term is your median. Now, your group of flowers looks like this:
1-12: 10 inches
13-17: 12 inches
18-21: 15 inches
The 11th item is going to be in that first group, meaning that the median is 10 inches.
If you prefer to write out all of the terms, it will be:
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 15, 15, 15, 15
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