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.
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$\begingroup$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 $6$.
$\endgroup$ $\begingroup$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
$$\sqrt{8-4\sqrt3}=2\cdot \sqrt{2-\sqrt3}= 2\left(\sqrt{\frac32} -\sqrt{\frac12}\right)=\sqrt6-\sqrt2 $$
$\endgroup$ $\begingroup$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.
$\sqrt{8-4\sqrt{3}} = \sqrt{6}-\sqrt{2}$
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