Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$

Question

How to find the square root of an irrational expression, to simplify that root. e.g.:

$$ \sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2} $$

Easy to verify:

\begin{align} (\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2 = 8 - 4 \sqrt{3} \end{align}

But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...

BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.

Answer 1

I don't think there's a name for this procedure but let's apply it to $\sqrt{8-4\sqrt3}$. If you suspect this equals $\sqrt a\pm\sqrt b$ with rationals $a$ and $b$, then $$8-4\sqrt3=(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt{ab}$$ so you want to solve simultaneously $a+b=8$ and $-4\sqrt{3}=\pm2\sqrt{ab}$. So you need the minus sign, and $ab=12$. Then $a$ and $b$ are roots of the quadratic equation $$(X-a)(X-b)=X^2-(a+b)X+ab=X^2-8X+12.$$ This does has rational roots: $2$ and $6$.

Answer 2

If I recall correctly, you make the assumption that your expression takes the form $\sqrt{a}\pm\sqrt{b}$:

$\sqrt{8-4\sqrt{3}} = \sqrt{a}\pm\sqrt{b}$

$8-4\sqrt{3} = 8-\sqrt{48} = a\pm2\sqrt{ab}+b$

We can see that the irrational part must be assigned the negative sign.

Equating rational and irrational parts:

$a+b = 8$

$-2\sqrt{ab} = -\sqrt{48} \implies ab = 12$

Then $a = \frac{12}{b} \implies \frac{12}{b}+b = 8 \implies 12 + b^2 = 8b \implies b^2 - 8b + 12 = (b-6)(b-2) = 0$

Since we know the answer is positive, take b = 2 and a = 6.

$\sqrt{8-4\sqrt{3}} = \sqrt{6}-\sqrt{2}$

Answer 3

Note the denesting formula

$$ \sqrt{a-\sqrt c} = \sqrt{\frac{a+\sqrt {a^2-c}}2}- \sqrt{\frac{a-\sqrt {a^2-c}}2} $$

which can be verified by squaring both sides, and apply it to

$$\sqrt{8-4\sqrt3}=2\cdot \sqrt{2-\sqrt3}= 2\left(\sqrt{\frac32} -\sqrt{\frac12}\right)=\sqrt6-\sqrt2 $$

Answer 4

I'm assuming you're looking for a formula to de-nest $\sqrt{a+b\sqrt{c}}$.

To derive such a formula, equate this to $x+y\sqrt{c}$.
$a+b\sqrt{c}=(x+y\sqrt{c})^2=x^2+cy^2+2xy\sqrt{c}$
$a+b\sqrt{c}=x^2+cy^2+2xy\sqrt{c}$

On matching up coefficients, we get:
$ \begin{cases} x^2+cy^2=a\\ 2xy=b \end{cases} $

Substitute $y=\frac{b}{2x}$:
$\rightarrow x^2+c(\frac{b^2}{4x^2})=a$
$u:=x^2\rightarrow u+\frac{cb^2}{4u}=a$
$\rightarrow u^2-au+\frac{b^2c}{4}=0$
$\rightarrow u=\frac{a\pm\sqrt{a^2-b^2c}}{2}$
$\because u=x^2\rightarrow x=\sqrt{\frac{a\pm\sqrt{a^2-b^2c}}{2}}$
Now solve for $y$:
$y=\frac{b}{2}(\sqrt{\frac{2}{a\pm\sqrt{a^2-b^2c}}})$

And you're done! This tells us that: $$\sqrt{a+b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2c}}{2}}+(\frac{b}{2}(\sqrt{\frac{2}{a+\sqrt{a^2-b^2c}}}))\sqrt{c}$$