Solution 3.1:3c
From Förberedande kurs i matematik 1
We start by looking at one part of the expression \displaystyle \sqrt{16}. This subexpression can be simplified since \displaystyle 16 = 4\cdot 4 = 4^{2} which gives that \displaystyle \sqrt{16} = \sqrt{4^{2}} = 4 and the whole expression becomes
\displaystyle \sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.} |
Can \displaystyle \sqrt{20} be simplified? In order to answer this, we split 20 up into integer factors,
\displaystyle 20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5 |
and see that 20 contains the square \displaystyle 2^2 as a factor and can therefore be taken outside the root sign,
\displaystyle \sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.} |