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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
avatar+1005 
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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
avatar+1005 
+1

you can also guess and check

proyaop  Oct 6, 2022

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