Challenge problem involving square roots

[lookagain]

without using the following method in the quote box:


\(\displaystyle \sqrt{ \ 2 + \sqrt{3} \ } \ + \ \sqrt{ \ 2 - \sqrt{3} \ } \)

\(\displaystyle x = \sqrt{ \ 2 + \sqrt{3} \ } \ + \ \sqrt{ \ 2 - \sqrt{3} \ } \)

Square both sides:

. . \(\displaystyle x^2 \;=\;\left[\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}\right]^2\)

. . \(\displaystyle x^2 \;=\;\left(\sqrt{2+\sqrt{3}}\right)^2 + 2\sqrt{(2+\sqrt{3})(2-\sqrt{3})} + \left(\sqrt{2-\sqrt{3}}\right)^2\)

. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2\sqrt{4-3} \:+\: 2-\sqrt{3}\)

. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2 \:+\: 2 - \sqrt{3}\)

. . \(\displaystyle x^2 \;=\;6\)


Therefore: .\(\displaystyle x \:=\:\sqrt{6}\)

 

[MarkFL]

\(\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\)

\(\displaystyle \sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}=\)

\(\displaystyle \sqrt{\dfrac{(\sqrt{3}+1)^2}{2}}+\sqrt{\dfrac{( \sqrt{3}-1)^2}{2}}=\)

\(\displaystyle \dfrac{\sqrt{2}}{2}\left(\sqrt{3}+1+\sqrt{3}-1 \right)=\dfrac{2\sqrt{2}\sqrt{3}}{2}=\sqrt{6}\)

 

[soroban]

Lovely solution, Mark!

 

[lookagain]

Please do not post portions of a problem within the title area. Thank you.

I was unsuccessful in changing the title to this thread.

 

[MarkFL]

I was unsuccessful in changing the title to this thread.

You probably also want to avoid embedding \(\displaystyle \LaTeX\) in topic titles in the future as well, as this increases server strain.

I recommend reporting the first post to the staff and request that the title be edited.