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!

Welcome to Math.SE. Please read how to ask a good question here: http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960 — M47145, Jul 18 '16 at 23:07
I can't think of anything besides possibly factoring out a $2$ — Colbi, Jul 18 '16 at 23:21
Take out the $\sqrt{2}$. The rest will be the square of something nice. — André Nicolas, Jul 18 '16 at 23:32
OK, call it $2$, it becomes $\sqrt{2}$. So now we are at $\sqrt{2}\sqrt{7+4\sqrt{3}}$. After the smoke clears you will end up with $\sqrt{2}(2+\sqrt{3})$, that is, $2\sqrt{2}+\sqrt{6}$. — André Nicolas, Jul 18 '16 at 23:37

score 11 · Answer 1 · answered Jul 19 '16 at 00:03

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 $a^2-b=4$.

score 3 · Answer 2 · answered Jul 18 '16 at 23:25

3

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 $ab=4$.

answered Jul 18 '16 at 23:25

Bernard

175,478

1

What if $a$ is a radical? – imranfat Jul 19 '16 at 01:49
Precisely, you find a radical. I never said $a$ and $b$ have to be integers. – Bernard Jul 19 '16 at 10:19

score 2 · Answer 3 · edited Apr 13 '17 at 12:20

2

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.

edited Apr 13 '17 at 12:20

Community

1

answered Jul 19 '16 at 01:47

Bill Dubuque

272,048

Thank you Bill I appreciate the new rule and will try to remember it, but can you clarify on what a norm and trace is? – Pikachu Jul 19 '16 at 13:56
@Pikachu Please see the post linked in the first sentence of the answer. – Bill Dubuque Jul 19 '16 at 14:02
Oh ok there it is... – Pikachu Jul 19 '16 at 23:30

score 1 · Answer 4 · answered Jul 19 '16 at 00:42

1

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.

answered Jul 19 '16 at 00:42

student forever

1,266

score 0 · Answer 5 · edited Sep 16 '19 at 03:25

First, we guess that $\sqrt{14 + 8 \sqrt{3}}$ can be written in the form $a + b$, where $a$ and $b$ are simple numbers that may involve square roots. (For example, if we square $1 + \sqrt{3}$, then we get $(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 4 + 2 \sqrt{3}$, so $\sqrt{4 + 2 \sqrt{3}} = 1 + \sqrt{3}$.)

So we have that $a + b = \sqrt{14 + 8 \sqrt{3}}.$ Squaring both sides, we get $a^2 + 2ab + b^2 = 14 + 8 \sqrt{3}.$ To solve for $a$ and $b$, we must obtain at least two equations, so we try setting $a^2 + b^2 = 14$ and $2ab = 8 \sqrt{3}$. Then $ab = 4 \sqrt{3}$. Squaring this equation, we get $a^2 b^2 = (4 \sqrt{3})^2 = 48$.

So we have the sum and product of $a^2$ and $b^2$. Then the quadratic whose roots are $a^2$ and $b^2$ is [(x - a^2)(x - b^2) = x^2 - (a^2 + b^2) x + a^2 b^2 = x^2 - 14x + 48 = 0.] This quadratic factors as $(x - 6)(x - 8) = 0$, so the roots are 6 and 8.

This tells us that $a^2$ and $b^2$ are equal to 6 and 8 in some order. Therefore, we suppose that \begin{align*} \sqrt{14+8\sqrt{3}} &= \pm \sqrt{6} \pm \sqrt{8} \\ &= \pm \sqrt{6} \pm 2\sqrt{2}. \end{align*}

We can easily see that $\sqrt{14+8\sqrt 3}$ is large enough that both signs above must be $+$. Thus $\sqrt{14 + 8 \sqrt{3}} = \boxed{2 \sqrt{2} + \sqrt{6}}.$

We can check this answer by squaring it: $(2 \sqrt{2} + \sqrt{6})^2 = 8 + 2 \cdot 2 \sqrt{2} \cdot \sqrt{6} + 6 = 14 + 8 \sqrt{3}.$

You missed a few dollar signs as a few Jax expressions are not rendering... — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Sep 14 '19 at 14:31

score 0 · Answer 6 · answered Jul 11 '20 at 19:26

0

Note

$$\sqrt {14 + 8\sqrt {3}}=\sqrt{2(7+2\sqrt{12})} =\sqrt{2(\sqrt4+\sqrt3)^2}=\sqrt2(2+\sqrt3) $$

answered Jul 11 '20 at 19:26

Quanto

97,352

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

6 Answers6

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