Fully simplify $\sqrt {14 + 8\sqrt {3}}$.

Question

Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.

I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$ So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.)

So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!

Answer 1

Hint:

Try to write $\;\sqrt{14+8\sqrt 3}=a+b\sqrt 3$. This means $$(a+b\sqrt3)^2=14+8\sqrt 3.$$ This relation will be satisfied if $\;\begin{cases}a^2+3b^2=14,\\ab=4.\end{cases}$

If $a^2$ and $b^2$ are integers, there are not so many possibilities to have $a^2+3b^2=14$. Try to check if any of these possibilities also satisfies $ab=4$.

Answer 2

Write $$ \sqrt{14+8\sqrt{3}}=\sqrt{x\vphantom{b}}+\sqrt{\smash{y}\vphantom{b}} $$ Then $$ 14+8\sqrt{3}=x+y+2\sqrt{\smash{xy}\vphantom{b}} $$ so we can look for $$ xy=48,\qquad x+y=14 $$ that reduces to finding the roots of the equation $$ z^2-14z+48=0 $$ that are $8$ and $6$. So we can take $x=8$ and $y=6$: $$ \sqrt{14+8\sqrt{3}}=\sqrt{6}+\sqrt{8} $$

Alternatively, recall the identity $$ \sqrt{a+\sqrt{b}}= \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}} $$ In this case $a^2-b=4$.

Answer 3

By Bernard's answer: $$\;\begin{cases}a^2+3b^2=14,\\ab=4.\end{cases}$$ So $$a^4+3(ab)^2=14a^2 \Rightarrow a^4-14a^2+3\cdot4^2=0\Rightarrow u^2-14u+6\cdot 8=0 \Rightarrow (u-6)(u-8)=0$$ where $u=a^2$. Now we can find all possibilities.

Answer 4

This can be computed by a very simple $ $ Square Root Denesting Rule:

Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

$\ 14+8\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\ 12+8\sqrt 3\:$

with $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{24}\, =\, 2\sqrt 6.\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of above yields $\,\ \sqrt 8 + \sqrt 6$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.