How does $(3\sqrt3)^2 = 27$?

I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$. it has been a while since I've done this kind of thing so it may be something obvious that I can't see.

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    $\begingroup$ FOIL? There is no binomial here. $9 + 3 \sqrt{3} + 3 \sqrt{3} + 3$ would be $(3 + \sqrt{3})^2$, not $(3 \sqrt{3})^2$. $\endgroup$ – Robert Israel Oct 14 '13 at 20:38
  • $\begingroup$ ah yes. basically expanding in the form (a•b)(a•b) edit: in fact that's (a+-b)(a+-b) too. my bad $\endgroup$ – mike Oct 14 '13 at 20:40
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    $\begingroup$ The total reputation of the people in this page is excessively too damn high! $\endgroup$ – Ali Oct 15 '13 at 3:34

Recall that $$(ab)^2 = (a)^2 (b)^2$$

Here, $a = 3$ and $b = \sqrt 3$.

$$(3 \sqrt 3)^2 = (3)^2(\sqrt 3)^2 = 9 \cdot 3 = 27.$$

  • $\begingroup$ I wonder if there is a 2.5x nice answer? :-) +1 $\endgroup$ – Amzoti Oct 16 '13 at 0:54

If you really insist on expanding, just use the fact that $3\sqrt{3} = \sqrt{3} + \sqrt{3} + \sqrt{3}$, and therefore expand away $(\sqrt{3} + \sqrt{3} + \sqrt{3})^2$ using the trinomial expansion formula $(a+b+c)^2 = a^2 + b^2 +c^2 +2ab+2bc+2ac$.

Later edit:

And if you really insist on using FOIL, then use a little imagination to write $3\sqrt{3} = 2\sqrt{3} + \sqrt{3}$, thus your desired expression becoming $(2\sqrt{3} + \sqrt{3})(2\sqrt{3} + \sqrt{3})$.


Your mistake is that $9+3\sqrt3+3\sqrt3+3=(3+\sqrt3)^2$. However $3\sqrt3\neq3+\sqrt3$.

As other answers indicate, $(3\sqrt3)^2=(\sqrt{3\cdot3^2})^2=(\sqrt{27})^2=27$.


Other people have basically already said it, but the FOIL is only something that you can apply when you have expressions of exactly this form: $$ (a + b)(c+d). $$ In this case you do: (F)irst (O)utside (I)nside (L)ast: The First part of $(a + b)^2$ is $ac$. The Outside is $ad$, the Inside if $bc$ and the Last is $bd$. In all you get $$ ac + ad + bc + bd. $$

Now consider what happens when you have $$ (a+b)^2 = (a+b)(a+b). $$ In this case you get $$ aa+ ab + ba + bb = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. $$

Now, you are asking about what to do with $$ (3\sqrt{3})^2 $$ Note that this is not of exactly the form $(a+b)(c+d)$ or $(a+b)^2$. You have the product of $3$ and $\sqrt{3}$ and to use FOIL you would need the sum $3+\sqrt{3}$. Therefore it doesn't make sense to use FOIL here. It simply isn't applicable (unless you rewrite the expression first as explain in one of the other answers).

So what do you do in this case? Your expression is exactly of the form $$ (ab)^2 $$ with $a=3$ and $b = \sqrt{3}$. And then we have a rule that says that $$ (ab)^2 = a^2b^2. $$ So you get $$ (3\sqrt{3})^2 = 3^2(\sqrt{3})^2 = 9\cdot 3 = 27. $$ And that is it.

You could also do it slightly differently by using the rule/definition $$ (ab)^2 = abab $$ So $$ (3\sqrt{3})^2 = 3\sqrt{3}\cdot3\sqrt{3} = 3\cdot 3\cdot \sqrt{3}\sqrt{3} = 9 \sqrt{3\cdot 3} = 9\sqrt{9} = 9\cdot 3 = 27. $$ Here we also used that the rule $\sqrt{c}\sqrt{d} = \sqrt{cd}$.

  • $\begingroup$ FOIL "simply isn't applicable"? How about writing $3\sqrt{3} = 2\sqrt{3} + \sqrt{3}$? $\endgroup$ – baudolino Oct 14 '13 at 21:09
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    $\begingroup$ @baudolino: Yes, I just saw your edit and upvoted your answer. One can of course rewrite the expression so that FOIL works, but as it is written $3\sqrt{3}$ you can't. You have to rewrite first. $\endgroup$ – Thomas Oct 14 '13 at 21:10
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    $\begingroup$ True. Although, in this particular instance, I wish mike would tell us if he struggles with the difference between $(a+b)^2$ and $(ab)^2$, or whether there is something more fundamental at work, such as not realizing that $(\sqrt{3})(\sqrt{3}) = 3$. $\endgroup$ – baudolino Oct 14 '13 at 21:16
  • $\begingroup$ @baudolino: I agree. $\endgroup$ – Thomas Oct 14 '13 at 21:17
  • $\begingroup$ The last part here is the best answer. $\endgroup$ – JP McCarthy Oct 16 '13 at 12:48

$(3\sqrt{3})^2=3^2\cdot\sqrt{3}^2=9\cdot 3=27$

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    $\begingroup$ sorry, I should've added this into my question. I do understand how this method works, but I just don't understand why FOIL or binomial expansion return something different $\endgroup$ – mike Oct 14 '13 at 20:37
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    $\begingroup$ @mike: FOIL only applies when you are multiplying two expressions, each with two terms. You are multiplying two expressions each with one term, with two factors. (The highlighted words have very specific distinct meanings in mathematics.) $\endgroup$ – DJohnM Oct 14 '13 at 20:41
  • $\begingroup$ lock colleague explained below, so I believe that there is no need to repeat the same to me $\endgroup$ – Madrit Zhaku Oct 14 '13 at 20:42

To use FOIL or binomial expansion, the term has to be like this: $$ (3 + \sqrt{3})^2 $$


This isn't an answer, but it's relevant and doesn't fit into a comment. You seem to have a misunderstanding of what FOIL is.

FOIL only works when you're multiplying $(a + b)(c + d)$. But why does it only work then? Likely, you were just taught it as a rule, with no explanation or justification for it, because high school doesn't do that sort of thing enough.

Think of it this way: FOIL isn't a new rule, it's just a way to think about the distributive law twice:

$$(a + b)(c + d) = (a + b)c + (a + b)d = (ac + bc) + (ad + bd) = ac + ad + bc + bd$$

Note how $ac$ is the firsts, $ad$ is the outers, $bc$ is the inners, and $bd$ is the lasts. Does it make more sense why you can't FOIL $(ab)\cdot(cd)$ now?






$\frac{x}{3^2}=3$ (Assuming $x>0$).




Here is a rule I think ...

  1. $(a\sqrt b)^2 = (\sqrt (a^2.b)^2$
  2. $(3\sqrt 3)^2 = (\sqrt (3^2.3)^2$
  3. $(3\sqrt 3)^2 = (\sqrt (9.3)^2$
  4. $(3\sqrt 3)^2 = (\sqrt (27)^2$
  5. $(3\sqrt 3)^2 $= 27

Henced proved....


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