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I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering: $$\sqrt{6}-\sqrt{2} = \sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}}=\sqrt{ 8-4\sqrt{3}}$$ But how can I do it the other way, also for other similar problems? I tried multiple problem solving sites, but none could simplify $\sqrt{8-4\sqrt{3}}$ further then $2\sqrt{2-\sqrt{3}}$

Shaun
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Boas
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    Just write it from right to left. The only place you may ellaborate is that $\sqrt{ (\sqrt{6}-\sqrt{2} )^{2}}=\sqrt{6}-\sqrt{2}$ because $\sqrt{6}-\sqrt{2}>0$ – Momo Apr 02 '22 at 16:01
  • This relies on noticing that $8-4\sqrt{3}=(\sqrt{6}-\sqrt{2})^2$. In general it is not easy, there is no oter solution than guessing and trying cases. – Marcos Apr 02 '22 at 16:03
  • Hint: note $\sqrt{8\pm4\sqrt{3}}$ have product $4$, so try the Ansatz $\sqrt{8\pm4\sqrt{3}}\sqrt{n+4}\pm\sqrt{n}$. Now take the difference of two squares. – J.G. Apr 02 '22 at 16:04
  • Alt. hint: if $x = \sqrt{8-4 \sqrt{3}}$ then squaring twice to eliminate the radicals gives $x^4 - 16 x^2 + 16 = 0$. Completing the squares in two ways gives the two factorizations $x^4 - 16 x^2 + 16$ $= (x^2 - 4 \sqrt{3} - 8) (x^2 + 4 \sqrt{3} - 8)$ $= (x^2 - 2 \sqrt{2}, x - 4) (x^2 + 2 \sqrt{2}, x - 4)$ which correspond to the two forms of the roots $,{,\pm \sqrt{8 \pm 4 \sqrt{3}},} = {,\pm \sqrt{2} \pm \sqrt{6},},$. – dxiv Apr 02 '22 at 17:06

2 Answers2

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If you didn't know the answer already for $\sqrt{8-4\sqrt{3}}$, you could start with $$8-4\sqrt{3}=(\sqrt{a}-\sqrt{b})^2$$ $$\implies8-4\sqrt{3}=a-2\sqrt{ab}+b$$ $$\implies a+b=8$$ and $$\sqrt{ab}=2\sqrt{3}\implies ab=12$$ and solve for $a$ and $b$

David Quinn
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  • I already thought of something like this, the problem basically is factorizing. You can use the same trick for quadratic formulas. One thing I don't understand is how you get the two numbers, the 8 and the 12. – Boas Apr 02 '22 at 16:14
  • by comparing rational and irrational parts after multiplying out the bracket – David Quinn Apr 02 '22 at 16:16
  • I don't fully understand you. So if you have $a-b\sqrt{c}=(\sqrt{p}-\sqrt{q})^2$ the two equations are $p+q=a$ and $p×q=b×c$? – Boas Apr 02 '22 at 16:21
  • please see my edited answer – David Quinn Apr 02 '22 at 16:25
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In the general case let solve: $$\sqrt{x\pm y}=\sqrt{a}\pm\sqrt{b}$$

Since LHS exsits and is positive we will also assume $x\ge y$ and $a\ge b$.

By squaring expressions we get the system $\begin{cases}x+y=a+b+2\sqrt{ab}\\x-y=a+b-2\sqrt{ab}\end{cases}\iff \begin{cases}x=a+b\\y=2\sqrt{ab}\end{cases}$

Solving for $a,b$ we get $\begin{cases}a=\frac 12 x+\frac 12\sqrt{x^2-y^2}\\b=\frac 12 x-\frac 12\sqrt{x^2-y^2}\end{cases}$

And there you have it:

$$\sqrt{\vphantom{|}x\pm y}=\sqrt{\frac 12 x+\frac 12\sqrt{x^2-y^2}}\pm\sqrt{\frac 12 x-\frac 12\sqrt{x^2-y^2}}$$

Applying for $x=8$ anf $y=4\sqrt{3}$ we get $\sqrt{x^2-y^2}=\sqrt{64-16\times 3}=\sqrt{16}=4$

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

zwim
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