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Math Help - the square of a sum

  1. #1
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    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?
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  2. #2
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    Re: the square of a sum

    Quote Originally Posted by hannah2329 View Post
    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.
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  3. #3
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    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.
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  4. #4
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    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?
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