Sqaure root of an expression with a square root

What do you mean by "finding" the square root? Do you mean "simplifying" in some manner? "Restating as a non-nested root"? "Evaluating" with a calculator?

Please reply with the full text of the exercise, including the instructions, and please show all of the steps you have tried so far.

Thank you.

Eliz.
 
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 \(\displaystyle 2\,+\,\sqrt{3}\)

\(\displaystyle \L\;\;\sqrt{2\,+\,\sqrt{3}}\;=\;\frac{\sqrt{2}\,+\,\sqrt{6}}{2},\;\frac{-\sqrt{2}\,-\,\sqrt{6}}{2}\)
.
 
soroban said:
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 \(\displaystyle \,a\,+\,b\sqrt{3}\;=\;\sqrt{2\,+\,\sqrt{3}}\)

I understand that theorem, but I do not see how the next step follows from it. How can you just remove the square root sign?
 
Hello, Trenters4325!

Let \(\displaystyle \,a\,+\,b\sqrt{3}\;=\;\sqrt{2\,+\,\sqrt{3}}\)

I understand that theorem, but I do not see how the next step follows from it.
How can you just remove the square root sign?
I'm puzzled by your question . . .

I said:
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: \(\displaystyle \,a^2\,+\,2ab\sqrt{3}\,+\,3b^2\;=\;2\,+\,\sqrt{3}\)

Which part of\(\displaystyle :\;Square\,\cdots\,both\,\cdots\,sides\,\) don't you understand?
 
I see. Sorry about that.

Also, I do know De Moivre's Theorem. Is there a simpler way using that?
 
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