Sqaure root of an expression with a square root

[Trenters4325]

How would I find the square root of the expresssion 2+sqrt(3)?

 

[stapel]

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.

 

[soroban]

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 $\displaystyle 2\,+\,\sqrt{3}$

\(\displaystyle \L\;\;\sqrt{2\,+\,\sqrt{3}}\;=\;\frac{\sqrt{2}\,+\,\sqrt{6}}{2},\;\frac{-\sqrt{2}\,-\,\sqrt{6}}{2}\)
.

 

[Trenters4325]

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?

 

[soroban]

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 . . .

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?

 

[Trenters4325]

I see. Sorry about that.

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