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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?

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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 $±\sqrt 3$).

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${\sqrt {7 + 4{\sqrt3}}} = {\sqrt {7 + \sqrt{48}}}$

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} $

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    $\begingroup$ That's a handy formula to remember for problems like this. It takes the guesswork out of it. (Although it does require that $a^2 > b$ if you want to keep it real.) $\endgroup$ – David K Jun 16 '16 at 13:28
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    $\begingroup$ @DavidK Aw, we getting real now! $\endgroup$ – Simply Beautiful Art Jun 16 '16 at 14:48
  • $\begingroup$ Or, as I write it, $$\sqrt{a+b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2c}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b^2c}}{2}}$$ $\endgroup$ – Mr Pie Apr 14 at 13:51
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$$\sqrt { 7+4\sqrt { 3 } } =\sqrt { 7+2\cdot 2\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } \right) }^{ 2 }+{ 2 }^{ 2 }+4\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } +2 \right) }^{ 2 } } =\sqrt { 3 } +2\\ \\ $$

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    $\begingroup$ I think you have may have a mistake in your third term there $\endgroup$ – KBusc Jun 16 '16 at 15:31
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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 $a^2+3b^2=7$

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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.

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    $\begingroup$ You dont have to look at it you can analytically solve this system :). As $b\neq 0$ we obtain $a=2\sqrt{3}/b$. Plug this into the other equation $12/b^2+b^2=7$ or $12+b^4=7b^2$. This can be solved as it is in essence only a quadratic equation, after the substitution $u=b^2$. $\endgroup$ – MrYouMath Jun 16 '16 at 22:25
  • $\begingroup$ Of course, you can solve it analytically. The point I was trying to make is that it is not difficult to solve and you can even see it directly. $\endgroup$ – Tom Jun 17 '16 at 7:58
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$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$

$a+\sqrt{b}=x+y+2\sqrt{xy}$

$a+\sqrt{b}=(x+y)+\sqrt{4xy}$

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!

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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. $$\sqrt[4]{27}+\sqrt[4]{12}$$

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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 $Y=2-\sqrt{3}$

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  • $\begingroup$ Wow! Nicely done! $(+1)$ :P $\endgroup$ – Mr Pie Apr 14 at 14:13

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