Having trouble simplyfing radicals of this sort

Question

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?

Answer 1

Starting with $$\sqrt{43+24\sqrt{3}}=4+3\sqrt{3}$$ you can square both sides to get $$ \begin{eqnarray} 43+24\sqrt{3} & = & (4+3\sqrt{3})^2 \\ & = & (4+3\sqrt{3})(4+3\sqrt{3}) \\ & = & 16+12\sqrt{3}+12\sqrt{3}+27 \\ & = & 43+24\sqrt{3} \\ \end{eqnarray} $$

Answer 2

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

Answer 3

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}

Answer 4

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 $b=3$.

Answer 5

Hint: Generally, you can solve such problems as : $$\sqrt { a\pm 2\sqrt { b } } =\sqrt { m } \pm \sqrt { n } $$ where is $\\ \\ \\ \begin{cases} a=m+n \\ b=mn \end{cases}$

Answer 6

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.