+308 \(\sqrt{4-2\sqrt{3}}\) 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!
+1632 \(a + b\sqrt{c} = \sqrt{4 - 2\sqrt{3}}\)
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...
