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Simplify the surd.

$(2\sqrt 3 + 3\sqrt 2)^2$

I know I should us this formula: $(a^2+2ab+b^2)$

But this gets complicated later. Please explain. :(

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Well, $a=2\sqrt{3}$ and $b=3\sqrt{2}$ and $(a+b)^2=a^2+2ab+b^2$. Now just plug it in and calculate. What is $a^2$? What is $b^2$? What is $2ab$? – Your Ad Here Jan 31 '14 at 18:34
The result will be of the form $x+y\sqrt 6$ – AlexR Jan 31 '14 at 18:34
For what it's worth, the calculations that others have presented become marginally easier if you first perform the admittedly non-obvious step of putting everything under a radical: $2\sqrt{3} + 3\sqrt{2} = \sqrt{2^2 \cdot 3} + \sqrt{3^2 \cdot 2} = \sqrt{12} + \sqrt{18}$ – Dave L. Renfro Feb 6 '14 at 20:37
up vote 3 down vote accepted

Applying the formula you get $$(2\sqrt3+3\sqrt2)^2 = 4\cdot3 + 2\cdot 2\sqrt3\cdot 3\sqrt 2 + 9\cdot 2 = 30 + 12\sqrt6$$

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Alternatively, with a little simplification:


we can find the result:

$$\begin{align}[\sqrt{2}(\sqrt{6}+3)]^2&=2(\sqrt{6}+3)^2\\ & =2(6+2\sqrt{6}\cdot3+3^2) \\ &=2(15+6\sqrt{6}) \\ &=30+12\sqrt{6} \end{align}$$

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thank you so much! – Kiara Jan 31 '14 at 19:02
@Kiara You're welcome! ;-) – user93957 Jan 31 '14 at 19:03
@Aðøbe though true, what's the "simplification"? You still need the binomial formula and you need a few more steps than with the direct approach... – AlexR Jan 31 '14 at 19:07
@AlexR I mean using the binomial theorem for $\sqrt 6+3$ is less difficult for a beginner than $2\sqrt 3+3\sqrt 2$. The primary goal of my post was to provide an alternative technique. – user93957 Jan 31 '14 at 19:09

We are given:

$$(2\sqrt 3 + 3\sqrt 2)^2$$


$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)$$

Multiply using techniques such as the FOIL method:

$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)=4\sqrt9+6\sqrt6+6\sqrt6+9\sqrt4$$



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thank you so much! – Kiara Jan 31 '14 at 19:05

Without any formula try to calculate: $$ (2\sqrt {3} + 3\sqrt{2})^2 = (2\sqrt {3} + 3\sqrt{2}) \cdot (2\sqrt {3} + 3\sqrt{2}) $$ But it seems like a lot computing, can not it be simplified a bit? Yes, it can. The answer is in the formula you wrote, because:

Edit after @Zado comment: $$ (a+b)^2=(a+b)\cdot (a+b)=a^2 + ab + ba +b^2 $$ But since multiplication ($\cdot$) is commutative that is $$ab=ba$$ we can easier the formula even a little bit more to the final form $$ (a+b)^2 = \ldots = a^2 + 2ab + b^2 $$

Now it is a bit easier to do the computing, because you can put $a = 2\sqrt{3}$ and $b = 3\sqrt{2}$ and then use the formula.

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It's important for OP to note that the formula simply comes from doing $(a+b)^2 = (a+b)\cdot(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$ and "memorizing" the result. You aren't doing anything different by going one way or another. – Zado Jan 31 '14 at 18:44

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