How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2  \;=\;  4+4\sqrt{3} + 3  \;=\;  7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
 A: $$\sqrt { 7+4\sqrt { 3 }  } =\sqrt { 7+2\cdot 2\sqrt { 3 }  } =\sqrt { { \left( \sqrt { 3 }  \right)  }^{ 2 }+{ 2 }^{ 2 }+4\sqrt { 3 }  } =\sqrt { { \left( \sqrt { 3 } +2 \right)  }^{ 2 } } =\sqrt { 3 } +2\\ \\ $$
A: $\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
$a+\sqrt{b}=x+y+2\sqrt{xy}$
$a+\sqrt{b}=(x+y)+\sqrt{4xy}$
Then:
$a=x+y$ and $b=4xy$
$a^2=x^2+y^2+2xy$
$a^2-b=(x-y)^2$
If $x>y$ then:
$\sqrt{a^2-b}=x-y$
Let's take $c=\sqrt{a^2-b}$ then $c=x-y$
$c=x-y$ and $a=x+y$ then:
$x=\frac{a+c}{2}$ and $y=\frac{a-c}{2}$
Then:
$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$
Now applly it for $\sqrt{7+4\sqrt{3}}$.
$\sqrt{7+4\sqrt{3}}=\sqrt{7+\sqrt{48}}$
$c=\sqrt{49-48}$
$c=1$
$\sqrt{7+\sqrt{48}}=\sqrt{\frac{8}{2}}+\sqrt{\frac{6}{2}}$
Then:
$\sqrt{7+4\sqrt{3}}=2+\sqrt{3}$
solved!
A: If $7+4\sqrt{3}$ is a perfect square this means $7=a^2+3b^2$ is the sum of two squares in $\Bbb{Z}[\sqrt{3}]$ and $4\sqrt{3}=2ab\sqrt{3}$. So the possibilities are
$$a=\pm 1,b=\pm 2\\a=\pm 2,b=\pm 1$$
Only the latter gives $a^2+3b^2=7$
A: what you usually do with such a square root is the following:
Look at the term under root sign and try to write it as a binomal
$a^2+2ab+b^2$.
So, in your case:
$7+4\sqrt{3} = a^2+2ab+b^2$
The square root has to be in the $2ab$ term, so we may assume that
$2ab = 4 \sqrt3$
and
$a^2+b^2 = 7$.
This is a system of two (nonlinear) equations with two variables and by carefully looking at them, you get $a=2$ and $b=\sqrt3$.
Hope this helps.
A: In general, there is a formula where given the expression $\sqrt{X\pm Y}$, one can rewrite that into $\sqrt{\frac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}$ where $X$ and $Y$ can be any number and $X>Y$.
In your case, we have $X=7$ and $Y=4\sqrt{3}$ because $7<4\sqrt{3}$. Plugging the values into the formula, we get: $$\sqrt{\frac {7+\sqrt{7^2-(\sqrt{48})^2}}{2}}+\sqrt{\frac {7-\sqrt{7^2-(\sqrt{48})^2}}{2}}\tag{1}$$
The radical $\sqrt{7^2-(\sqrt{48})^2}$ simplifies into $1$ so $(1)$ becomes $$\sqrt{\frac {7+1}{2}}+\sqrt{\frac {7-1}{2}}$$ which further simplifies into $$\sqrt{4}+\sqrt{3}\iff\boxed{\sqrt{3}+2}$$

Extra: This works on any nested radical under a square root. For example: Denest $\sqrt{5\sqrt{3}+6\sqrt{2}}$. Setting $X=5\sqrt{3}$ and $Y=6\sqrt{2}$, gives us $$\sqrt{\frac {5\sqrt{3}+\sqrt{3}}{2}}+\sqrt{\frac {5\sqrt{3}-\sqrt{3}}{2}}\tag{1}$$
Simplifying gives us our denesting. $$\sqrt[4]{27}+\sqrt[4]{12}$$
A: If you want $\sqrt{7+4\sqrt{3}}=a+b\sqrt d$ (where $a$ and $b$ are rational numbers, and $d$ is a square-free integer) then $7+4\sqrt{3} = a^2+db^2+2ab\sqrt d$, which yields the natural choice $d=3$.
Then $a^2+3b^2=7$ and $2ab=4$ can be solved in the rational numbers, and you find $a=2,b=1$ (because $a^2+3 \cdot (2/a)^2=7$ has solutions $±2$ and $±\sqrt 3$).
A: ${\sqrt {7 + 4{\sqrt3}}} = {\sqrt {7 + \sqrt{48}}}$ 
The latter can be solved by the formula:
$\sqrt {a + \sqrt{b} } = \sqrt{ {a + \sqrt{a^2 -b}}\over2} +\sqrt{ {a - \sqrt{a^2 -b}}\over2} $
A: A very funny way to solve this which can be generalized goes as follow:
We define $X=\sqrt{7+4\sqrt{3}}$ and $Y= \sqrt{7-4\sqrt{3}}$. Note that $X\pm Y>0$. Now we form $X+Y$ and $X-Y$

$(X+Y)^2=X^2+Y^2+2XY=7+4\sqrt{3}+7-4\sqrt{3}+2\sqrt{49-48}=16\Longrightarrow X+Y=4$
$(X-Y)^2=X^2+Y^2-2XY=7+4\sqrt{3}+7-4\sqrt{3}-2\sqrt{49-48}=12\Longrightarrow X-Y=2\sqrt{3}$ 
  The last step is to solve the system:$$\begin{cases} 
X+Y=4\\
X-Y=2\sqrt{3}\\ 
\end{cases}$$
$X=2+\sqrt{3}$ and $Y=2-\sqrt{3}$

A: Use conjugates; start from $(a\pm b\sqrt{3})^2=7\pm 4\sqrt{3}$. Multiplying these equations gives $a^2-3b^2=\pm1$, whereas averaging them gives $a^2+3b^2=7$. If we want a rational $b$, the $\pm1$ should be $1$ so $b=1\implies a=2$.
