\(\sqrt{28 + 16\sqrt{3}}\)
If you want to turn that to \(a+ b\sqrt{c}\), here's how.
First the c is obviously a 3, because what else would it be duhhh.
\(28 + 16\sqrt{3} = (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2\)
\(2ab\sqrt{3} = 16\sqrt{3} \)
\(ab = 8\)
\(a^2 + 3b^2 = 28\)
If a and b are integers, you can guess and check a little bit. After some testing, you will find that:
\((a, b) = (4, 2)\)
So: \(\sqrt{28+16\sqrt{3}} = 4 + 2\sqrt{3}\)
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