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$$\sqrt{4-2\sqrt{3}}$$ can be expressed as $$a + b\sqrt{c}$$, where $$a$$, $$b$$, and $$c$$ are intergers and $$c$$ is squarefree. Find $$a + b + c$$.

Any help is appreciated!

Oct 2, 2022

#1
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$$a + b\sqrt{c} = \sqrt{4 - 2\sqrt{3}}$$

We can easily speculate the c = 3 (what else could c be equal to, 17!?).

$$a + b\sqrt{3} = \sqrt{4 - 2\sqrt{3}}$$

We can safely square both sides of the equation...

$$(a + b\sqrt{3})^2 = 4 - 2\sqrt{3}$$

Open up parenthesis: $$a^2 + 2ab\sqrt{3} + 3b^2 = 4 - 2\sqrt{3}$$

Then we can see that the root 3 part corresponds to the root 3 part in the equation since a and b are integers:

$$2ab\sqrt{3} = -2\sqrt{3}$$

$$ab = -1$$

From earlier, we also obtained:

$$a^2 + 3b^2 = 4$$

How can you take it from here to find a and b? Good luck...

Oct 3, 2022
#2
+1315
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you can also guess and check

proyaop  Oct 6, 2022