<PRE><B>Question 702409</B>
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There is one right next to the lead coefficient 4 and another right next to the -9 coefficient on the 1st degree term.


Oh!  You meant, &quot;how do I find the &lt;i&gt;&lt;b&gt;value&lt;/b&gt;&lt;/i&gt; of *[tex \LARGE x],&quot; didn&#39;t you?  You really should take the time to say what you mean so that we don&#39;t have to guess.


Let *[tex \LARGE u\ =\ x^{\small\frac{1}{4}}\LARGE].  Then *[tex \LARGE u^2\ =\ x^{\small\frac{1}{2}}\LARGE]. 



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4u^2\ -\ 9u\ +\ 4\ =\ 0]


Use the quadratic formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \frac{9\ \pm\ \sqrt{17}}{8}]. 


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ u^4\ =\ \left(\frac{9\ \pm\ \sqrt{17}}{8}\right)^4].


You can go ahead and do all of the ugly arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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