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      <text>10.1.3.1 Solutions to polynomial equations</text>
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      <text><![CDATA[<h3>Solutions to polynomial equations </h3>

<p>With the introduction of complex numbers we can now obtain solutions to those polynomial equations which may have real solutions, complex solutions or a combination of real and complex solutions. For example, the simple quadratic equation <span class="math inline">\(x^2+16=0\) </span>can be rearranged as <span class="math inline">\(x^2=-16\)</span> and then taking square roots we see that <span class="math inline">\(x=\pm\sqrt{-16}=\pm4\sqrt{-1}=\pm 4i\)</span> where we are replacing <span class="math inline">\(\sqrt{-1}\)</span> by the symbol ‘<span class="math inline">\(i\)</span>’.</p>

<p>This approach can be extended to the general quadratic equation <span class="math inline">\(ax^2+bx+c=0\) with roots \(\displaystyle{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\)</span> so that if for example <span class="math inline">\(3x^2+2x+2=0\)</span> then solving for <span class="math inline">\(x\)</span>: <span class="math display"><br></span></p><p><span class="math display">\[\begin{aligned}
x&amp;={-2\pm \sqrt{4-4(3)(2)}\over
{2(3)}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &amp;={-2\pm \sqrt{-20}\over
{6}}=  {-2\pm i\sqrt{20}\over
{6}}
\\\end{aligned}\]</span> <br></p><p>so, \(\large(\)as <span class="math inline">\(\displaystyle{\frac{\sqrt{20}}{6}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}}\)</span>\(\large)\), the two roots are <span class="math inline">\(\displaystyle{-{\frac13}
+\frac{\sqrt{5}}{3}i}\)</span> and <span class="math inline"></span><span>\(\displaystyle{-{\frac13} -\frac{\sqrt{5}}{3}i}\)</span>.</p><p>
In this example we see that the two solutions (roots) are <span><strong>complex conjugates</strong></span> of each other. In fact this will always be the case if the polynomial equation has coefficients: that is, if any complex roots occur they will always occur in complex conjugate pairs.</p>]]></text>
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