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## 'Ab Surd Ity' printed from http://nrich.maths.org/

Congratulations to Hyeyoun from St Paul's Girls School, London, to
Sue Liu of Madras College, St Andrew's, Scotland, Sanjay from The
Perse School, Cambridge and Bill from Alcester Grammar School for
your solutions.

We take the square root symbol in the question to signify the
positive square root. The tactic here is to square both sides and
then find the correct square root. If
$\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=X$, then $$X^2=
\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)
\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right).$$ The right hand
side equals $$2 + \sqrt{3}-2\left(\sqrt{2+\sqrt{3}}\times
\sqrt{2-\sqrt{3}}\right) +2-\sqrt{3},$$ and
$$\sqrt{2+\sqrt{3}}\times\sqrt{2-\sqrt{3}}=1.$$ Therefore $$X^2 =
2+\sqrt{3}-2+2-\sqrt{3}=2.$$ Does $X=-\sqrt{2}$ or
$+\sqrt{2}$?

Well $2+\sqrt{3}> 2-\sqrt{3}$, so $\sqrt{2+\sqrt{3}}>
\sqrt{2-\sqrt{3}}$, so $X$ is positive and we have
$X=\sqrt{2}$.

Note that we could take each square root to be positive or
negative. If so, then the question is much harder and there are
more solutions for $X$. For example, we could take $\sqrt{3} =
1{\cdot}732\cdots$; then $\sqrt{2+\sqrt{3}}$ has two values
(approximately $\pm 1{\cdot}93$), and $\sqrt{2-\sqrt{3}}$ has two
values (approximately $\pm 0{\cdot}52$). It follows that $X$ has
four values (approximately $\pm 2{\cdot}45$ and $\pm 1{\cdot}41$).
Alternatively, we could take $\sqrt{3} = -1{\cdot}732\cdots$, and
then we would get even more solutions.

In the second part there are again many solutions (because square
roots have two values and cube roots have three values). To
simplify the solution we restrict ourselves to real cube roots. We
want to find $$X = {\root 3\of {2+\sqrt{5}}} + {\root 3\of
{2-\sqrt{5}}}.$$ One way to do this is to write $a = {\root 3\of
{2+\sqrt{5}}}$ and $b= {\root 3\of {2-\sqrt{5}}}$, and use the
equation $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab(a + b)
+ b^3.$$ As $X = a + b$, we have $$X^3 = (2 + \sqrt{5}) + 3X{\root
3\of {(2 + \sqrt{5})(2 - \sqrt{5})}} +(2 - \sqrt{5}).$$ As $\root
3\of {(2 + \sqrt{5})(2 - \sqrt {5}} = \root 3\of {-1} = -1$ this
gives $X^3 + 3X - 4 = 0$ and hence $$(X - 1)(X^2 + X + 4) = 0.$$ As
$X^2 + X + 4 = 0$ has only complex solutions, and we are looking
for the real values of $X$, so we have $X = 1$, that is $${\root
3\of {2 + \sqrt{5}}} + {\root 3\of {2 - \sqrt{5}}} = 1.$$ You can
check this on your calculator!