<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>4.0.3 (Surds and radicals)</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Surds and radicals</h3>

<p>A <b>radical</b> is a number which is written with the symbol '\(\sqrt{}\)'. A <b>surd</b> is a real, irrational radical. For example, \(\sqrt{2}\) is a radical and a surd however \(\sqrt{4}\) is not a surd because it simplifies to \(2\). Remember that we define \(\sqrt{a}\) to be the non-negative number such that \(\sqrt{a} \cdot \sqrt{a} = a\). The following are some of the key properties of such numbers:</p>
<br/>
<ol type= "a">
<li>\(\sqrt{a} \geq 0 \: \forall a \in \mathbb{R}\).</li>
<li>\(\sqrt{a} \in \mathbb{R} \iff a \geq 0\).</li>
<li>\(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) \(\forall a,b \geq 0\).</li>
<li>\(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\) for \(a \geq 0\) and \(b > 0\).</li>
</ol>

<p>In addition to these properties we can also combine surds algebraically, using the operations of addition, subtraction and multiplication. However, division works slightly differently. If we have an expression of the form \(\frac{a}{\sqrt{b}}\) it is usual to remove the radical in the denominator by multiplying the whole fraction by \(\frac{\sqrt{a}}{\sqrt{a}}\). Since \(\frac{\sqrt{a}}{\sqrt{a}}=1\) performing this multiplication does not change the value of the fraction. This process is called <i>rationalising the denominator</i>.</p>
</br>
<h3>Examples</h3>

<p>Simplify the following expressions:</p>
<ol type="a">
<li>\((5-\sqrt{2})^2\).</li>
<li>\(\frac{5}{3-\sqrt{2}}\).</li>
</ol>

<h3>Solutions</h3>

<ol type="a">
<li>Note that \((5-\sqrt{2})^2 = (5-\sqrt{2})(5-\sqrt{2})^2\), thus we can expand the brackets to obtain
\[
\begin{align*}
(5-\sqrt{2})^2 &= (5-\sqrt{2})(5-\sqrt{2})
\\
&= 5^2 - 5 \sqrt{2} - 5 \sqrt{2} + (\sqrt{2})^2
\\
&= 25 - 10 \sqrt{2} + 2
\\
&= 27 - 10 \sqrt{2}.
\end{align*}
\]
</li>
<li>In this case, we wish to rationalise the denominator, so multiplying the fraction by \(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}\) we obtain
\[
\begin{align*}
\frac{5}{3-\sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} &= \frac{5 (3-\sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})}
\\
&= \frac{5 (3-\sqrt{2})}{3^2 - (\sqrt{2})^2}
\\
&= \frac{15 + 5\sqrt{2}}{9 - 2}
\\
&=  \frac{15 + 5\sqrt{2}}{7}.
\end{align*}
\]
</li>
</ol>

<p>Note that in part b. in order to rationalise the denominator we had to multiply the given fraction by \(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}\). We can generalise this process to by noting that for any fraction of the form \(\frac{c}{a+\sqrt{b}}\) we can remove the surd in the denominator by multiplying by \(\frac{a-\sqrt{b}}{a-\sqrt{b}}\). The expression \(a-\sqrt{b}\) is called the <b>radical conjugate</b> of \(a+\sqrt{b}\). We say that multiplying the top and bottom of the given fraction by the radical conjugate of the denomiator removes the surd in the denominator of the fraction.</p>]]></text>
<file name="2-1-fig-1-function-as-mapping.svg" path="/" encoding="base64">[encoded data, 307268 characters]</file>
<file name="figure-1-1.svg" path="/" encoding="base64">[encoded data, 307484 characters]</file>
    </questiontext>
    <generalfeedback format="html">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
