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.