Challenge problem involving square roots

lookagain

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without using the following method in the quote box:


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

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

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: .\(\displaystyle x \:=\:\sqrt{6}\)
 
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\(\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\)

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

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

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