From Förberedande kurs i matematik 1

Revision as of 09:10, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.1:3c moved to Solution 3.1:3c: Robot: moved page) ← Previous diff		Current revision (10:24, 30 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We start by looking at one part of the expression <math>\sqrt{16}</math>. This subexpression can be simplified since <math>16 = 4\cdot 4 = 4^{2}</math> which gives that <math>\sqrt{16} = \sqrt{4^{2}} = 4</math> and the whole expression becomes
-	<center> [[Image:3_1_3c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.}</math>}}
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		+	Can <math>\sqrt{20}</math> be simplified? In order to answer this, we split 20 up into integer factors,
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		+	{{Displayed math||<math>20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5</math>}}
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		+	and see that 20 contains the square <math>2^2</math> as a factor and can therefore be taken outside the root sign,
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		+	{{Displayed math||<math>\sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.}</math>}}

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Current revision

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{.}