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

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.

• The term of art (and it really is something of an art) is "denesting radicals". That should help you search for references. – Blue May 24 '20 at 5:36
• perhaps you could assume that the inner expression is a perfect square, then reform so that the inner radical has a coefficient of two. e.g. $\sqrt{8-4\sqrt{3}} = \sqrt{8-2\sqrt{12}}=\sqrt{6-2\sqrt{12}+2}=\dots$ – John Joy May 24 '20 at 6:12

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 $$6$$.

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

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

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.

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