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I'm studying radicals and rational exponents. I'm having lots of hardships with problems of this sort: prove $$\sqrt{43+24\sqrt{3}}=4+3\sqrt{3}$$ I keep going around and around experimenting with factoring. I can't seem to be able to prove this one in particular. Am I missing any common practice in regards to solving these and thus complicating it further? Is there any thing in particular I should always have in mind, or is this just lack of practice?

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    $\begingroup$ Just square both sides & verify they give the same thing ? $\endgroup$ – Donald Splutterwit Jul 3 '18 at 21:05
  • $\begingroup$ What happens when you multiply $4+3\sqrt{3}$ by itself? $\endgroup$ – AakashM Jul 4 '18 at 7:52
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We aim to complete the square under the radical sign, treating the first term (43) as the sum of squares $a^2+b^2$ and the second term ($24\sqrt{3}$) as $2ab$.

\begin{eqnarray}\sqrt{43+24\sqrt{3}} &=&\sqrt{43+2\cdot 4\cdot 3\sqrt{3}} \\&=&\sqrt{16+2\cdot 4\cdot 3\sqrt{3}+27}\\ &=&\sqrt{(4+ 3\sqrt{3})^2}\\&=&4+ 3\sqrt{3} \end{eqnarray}

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  • $\begingroup$ This is a brilliant answer from Angle, Daniel. It helps you achieve what wanted, which is to UNDERSTAND how to DERIVE the right hand side from the left. $\endgroup$ – Eureka Jul 5 '18 at 7:29
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HINT: Square both sides $$ (\sqrt{43 + 24 \sqrt{3}})^2 = 16 + 9(3) + 24 \sqrt{3},$$ $$ 43 + 24 \sqrt{3} = 43 + 24\sqrt{3}.$$

NOTE $(a+b)^2 = a^2 + 2ab + b^2$

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  • $\begingroup$ Assumption of what in fact, is to be proven. $\endgroup$ – Namaste Jul 3 '18 at 21:28
  • $\begingroup$ Prove that $-2 = 2$. "Hint: Square both sides to get $4 = 4$." Now, have you proven that $-2 = 2$? $\endgroup$ – Namaste Jul 3 '18 at 21:34
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    $\begingroup$ 1. You can reverse the argument order by yourself. 2. Clearly both sides are positive. $\endgroup$ – tonychow0929 Jul 4 '18 at 0:23
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    $\begingroup$ @amWhy A hint is not a proof, and not intended to be blindly followed when writing a proof. The idea of squaring both sides leads one to realize how to prove it, so it is a good hint. $\endgroup$ – JiK Jul 4 '18 at 12:44
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Starting with $$\sqrt{43+24\sqrt{3}}=4+3\sqrt{3}$$ you can square both sides to get $$ \begin{eqnarray} 43+24\sqrt{3} & = & (4+3\sqrt{3})^2 \\ & = & (4+3\sqrt{3})(4+3\sqrt{3}) \\ & = & 16+12\sqrt{3}+12\sqrt{3}+27 \\ & = & 43+24\sqrt{3} \\ \end{eqnarray} $$

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    $\begingroup$ Uummmm the question given to the OP, and stated by them, it is not "given" that $\sqrt{43+24\sqrt{3}}=4+3\sqrt{3}$ do something; rather, your assumption is precisely what is to be proven. $\endgroup$ – Namaste Jul 3 '18 at 21:27
  • $\begingroup$ Prove that $-2 = 2$. "Given $-2=2$, you can square both sides to get $(-2)^2 = 2^2 \iff 4 = 4$." Now, have you proven that $-2 = 2$? $\endgroup$ – Namaste Jul 3 '18 at 21:33
  • $\begingroup$ @amWhy, I don't mean to say that you are given that it is true, maybe I sure say "let us show that..." Also, I agree with you, and you are of course correct, that squaring a negative would not be proof. However, there is not a negative here. $\endgroup$ – Carser Jul 3 '18 at 21:42
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    $\begingroup$ @amWhy When you use equivalences, it doesn't matter which way you go, as all steps can be reversed to get a valid reasoning in the opposite direction. The answer lacks just one notice, that both sides of the claimed equality are positive, hence they are actually equal iff their squares are equal. With this (quite obvious) note the solution is valid. $\endgroup$ – CiaPan Jul 4 '18 at 7:50
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Hint: Generally, you can solve such problems as : $$\sqrt { a\pm 2\sqrt { b } } =\sqrt { m } \pm \sqrt { n } $$ where is $\\ \\ \\ \begin{cases} a=m+n \\ b=mn \end{cases}$

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The OP wants a strict proof of the statement, rather than how to derive it. In order to prove an equation, we need to either 1) go from one side to another, or 2) start with a trivially true statement (equation) and proceed to the claim. I chose the latter:

$$4+3\sqrt{3}=4+3\sqrt{3}\\ (4+3\sqrt{3})^2=(4+3\sqrt{3})^2\\ (4+3\sqrt{3})^2=43+24\sqrt{3}\\ 4+3\sqrt{3}=\sqrt{43+24\sqrt{3}} $$

Hence, the claim.

Notice that we choose only positive root, since we know $4+3\sqrt{3}$ is a positive number.

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  • $\begingroup$ This is a clear answer tracking useful priniciples that can be applied generally. Thank you. $\endgroup$ – Eureka Jul 5 '18 at 7:32
  • $\begingroup$ *showing useful principles... $\endgroup$ – Eureka Jul 5 '18 at 7:33
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One way is to solve $(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt{3}=43+24\sqrt{3}$ for integers $a$ and $b$. That is, $a^2+3b^2=43$, and $ab=12$. It is an easy exercise to solve the afore-mentioned system to get $a=4$ and $b=3$.

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