# the square of a sum

• Sep 10th 2011, 09:57 AM
hannah2329
the square of a sum
the problem is (c+3/c)^2 I'm trying to use the rules so i squared the first term c^2 then this is where i get stuck, double the product of the two terms, its 6 right? then square the last term which would be 9/c^2 do the c^2 cancel out to make the answer 6+9? i tried 15 as my answer on the online system and it was wrong. i only have a few more tries left and I'm not sure what else i could do to solve the problem?
• Sep 10th 2011, 10:13 AM
Prove It
Re: the square of a sum
Quote:

Originally Posted by hannah2329
the problem is (c+3/c)^2 I'm trying to use the rules so i squared the first term c^2 then this is where i get stuck, double the product of the two terms, its 6 right? then square the last term which would be 9/c^2 do the c^2 cancel out to make the answer 6+9? i tried 15 as my answer on the online system and it was wrong. i only have a few more tries left and I'm not sure what else i could do to solve the problem?

No, the c^2 does not cancel.
• Sep 10th 2011, 08:50 PM
scherz0
Re: the square of a sum
Hello there,

I assume that the expression is:

$\left(c + \frac{3}{c}\right)^2$.

As you said correctly, square the first term:

$c^2$.

Next, double the product of the two terms:

$2 \times c \times \frac{3}{c}$

Finally, square the last term to get:

$\frac{c^2}{9}$.

Notice that when you add only the first and third term above together, you get:
(I ignore the second term just so you can see the expression below)

$c^2 + \frac{1}{9}c^2$.

Do you see anything that can be simplified? In other words, does anything actually cancel?

I hope that this helps.
• Sep 11th 2011, 05:14 AM
HallsofIvy
Re: the square of a sum
In general, $(a+ b)^2= a^2+ 2ab+ b^2$. Here, a= c and $b= \frac{3}{c}$ so $a^2= c^2$ and $b^2= \frac{9}{c^2}$. What is 2ab?