The sum of two real numbers is 13, and the average of their squares is
1 more than the square of their average.  What is the difference of
their squares? (show work)

Hello.

The sum of two real numbers is 13:
x + y = 13

The average of their squares is:
(x^2 + y^2)/2 

The square of their average is:
((x+y)/2)^2

So:
(x^2 + y^2)/2 = ((x+y)/2)^2 + 1

                x + y = 13, so:

(x^2 + y^2)/2 = (13/2)^2 + 1

(x^2 + y^2)/2 = 43.25

(x^2 + y^2)/2 - 43.25 = 0

Multiplying each side by 2:

x^2 + y^2 - 86.5 = 0

Remember, x + y = 13, or x = 13 - y,   so:

(13 - y)^2 +  y^2 - 86.5 = 0

169 - 26y + 2y^2 - 86.5 = 0

Dividing by 2:

y^2 - 13y + 41.25 = 0

or:

(y - 5.5)(y - 7.5) = 0

So either y - 5.5 = 0, or y - 7.5 = 0

Thus,

y= 5.5 or 7.5

x + y = 13, so x= 5.5 or 7.5

Notice, that the original statement holds true: the average of their
squares is 1 more than the square of their average.
(((5.5)^2 + (7.5)^2)/2) - (((5.5 + 7.5)/2)^2) = 1
://www.google.com/search?hl=en&ie=ISO-8859-1&oe=ISO-8859-1&q=%28%28%285.5%29%5E2+%2B+%287.5%29%5E2%29%2F2%29+-+%28%28%285.5+%2B+7.5%29%2F2%29%5E2%29&btnG=Google+Search

The difference of their squares is:

(7.5^2) - (5.5^2) = 26
or
(5.5^2) - (7.5^2) = -26

---------

strategy:
algebra

I hope this helps.