Hello, Trenters4325!
Assuming you're not familiar with DeMoivre's Theorem,
\(\displaystyle \l\l\)here's an elementary method . . .
How would I find: \(\displaystyle \,\sqrt{2 + \sqrt{3}}\) ?
Theorem:
If we have: \(\displaystyle p\,+\,q\sqrt{3} \:= \:r\,+\,s\sqrt{3}\), then: \(\displaystyle \,p\,=\,r\,\) and \(\displaystyle \,q\,=\,s\)
\(\displaystyle \;\;\)That is, the corresponding coefficients are equal.
Let \(\displaystyle \,a\,+\,b\sqrt{3}\;=\;\sqrt{2\,+\,\sqrt{3}}\)
Square both sides: \(\displaystyle \,(a\,+\,b\sqrt{3})^2\;=\;\left(\sqrt{2\,+\,sqrt{3}}\right)^2\)
\(\displaystyle \;\;\)then: \(\displaystyle \,a^2\,+\,2ab\sqrt{3}\,+\,3b^2\;=\;2\,+\,\sqrt{3}\)
Since we have: \(\displaystyle \,\left(a^2\,+\,3b^2\right)\,+\,(2ab)\sqrt{3}\;=\;2\,+\,\sqrt{3}\)
\(\displaystyle \;\;\)then: \(\displaystyle \,\begin{array}{cc}a^2\,+\,3b^2 \:=\:2\\ 2ab\:=\:1\end{array}\)
Solve the system of equations . . .
Solve the second equation for \(\displaystyle b:\;\;2ab\:=\:1\;\;\Rightarrow\;\;b\:=\:\frac{1}{2a}\;\)
[1]
Substitute into the first equation: \(\displaystyle \;a^2\,+\,3\left(\frac{1}{2a}\right)^2\:=\:2\;\;\Rightarrow\;\;a^2\,+\,\frac{3}{4a^2}\:=\:2\)
Multiply by \(\displaystyle 4a^2:\;\;4a^4\,+\,3\:=\:8a^2\)
We have a "quadratic": \(\displaystyle \,4a^4\,-\,8a^2\,+\,3\;=\;0\)
\(\displaystyle \;\;\)which factors: \(\displaystyle \,(2a^2\,-\,1)(2a^2\,-\,3)\;=\;0\)
and has two equations to solve:
\(\displaystyle \;\;2a^2\,-\,1\:=\:0\;\;\Rightarrow\;\;a^2\:=\:\frac{1}{2}\;\;\Rightarrow\;\;a\:=\:\pm\frac{\sqrt{2}}{2}\)
\(\displaystyle \;\;2a^3\,-\,3\:=\:0\;\;\Rightarrow\;\;a^2\:=\:\frac{3}{2}\;\;\Rightarrow\;\;a\:=\:\pm\frac{\sqrt{6}}{2}\)
Substitute into
[1] and get the corresponding b-values: \(\displaystyle \,b\:=\:\pm\frac{\sqrt{2}}{2},\;\pm\frac{\sqrt{6}}{6}\)
It seems that we have four possible answers:
\(\displaystyle \;\;\frac{\sqrt{2}}{2}\,+\,\frac{\sqrt{2}}{2}\sqrt{3}\:=\:\frac{\sqrt{2}\,+\,\sqrt{6}}{2}\)
\(\displaystyle \;\;\)-\(\displaystyle \frac{\sqrt{2}}{2}\,-\,\frac{\sqrt{2}}{2}\sqrt{3}\:=\:\frac{-\sqrt{2}\,-\,\sqrt{6}}{2}\)
\(\displaystyle \;\;\frac{\sqrt{6}}{2}\,+\,\frac{\sqrt{6}}{6}\sqrt{3}\:=\:\frac{\sqrt{6}}{2}\,+\,\frac{\sqrt{18}}{6}\:=\:\frac{\sqrt{6}}{2}\,+\,\frac{3\sqrt{2}}{6}\:=\:\frac {\sqrt{6}}{2}\,+\,\frac{\sqrt{2}}{2}\)
\(\displaystyle \;\;\)-\(\displaystyle \frac{\sqrt{6}}{2}\,-\,\frac{\sqrt{6}}{6}\sqrt{3}\:=\:\)-\(\displaystyle \frac{\sqrt{6}}{2}\,-\,\frac{\sqrt{18}}{6}\:=\:\)-\(\displaystyle \frac{\sqrt{6}}{2}\,-\,\frac{3\sqrt{2}}{6}\:=\:\)-\(\displaystyle \frac{\sqrt{6}}{2}\,-\,\frac{\sqrt{2}}{2}\)
But we see that the last two are the same as the first two.
So, as expected, there are
two square roots of \(\displaystyle 2\,+\,\sqrt{3}\)
\(\displaystyle \L\;\;\sqrt{2\,+\,\sqrt{3}}\;=\;\frac{\sqrt{2}\,+\,\sqrt{6}}{2},\;\frac{-\sqrt{2}\,-\,\sqrt{6}}{2}\)
.