Question 801875
<font face="Times New Roman" size="+2">


This is what I have so far:
sqrt(2x-1)=sqrt(x-1)+1
2x+1=((sqrt(x-1) +1))((sqrt(x-1)+1)
2x-1=x-1+2sqrt(x-1)+1

Sign error right here:
(x<u>+</u>1)^2=(2sqrt(x-1))^2


(x+1)(x+1)=4(x-1)
x^2+2x+1=4x-4
x^2-2x+5=0


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2x\ -\ 1}\ -\ \sqrt{x\ -\ 1}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 1\ =\ x\ -\ 1\ +\ 2\sqrt{x\ -\ 1}\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 1\ =\ x\ +\ 2\sqrt{x\ -\ 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 1\ =\ 2\sqrt{x\ -\ 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ +\ 1\ =\ 4x\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 6x\ +\ 5\ =\ 0]


That should factor quite tidily.  However, don't depend on everything to factor neatly.  Some quadratics have irrational zeros and others have complex zeros.  Also, whenever you square both sides of an equation in the process of solving it, you must consider the possibility that you have introduced extraneous roots.  CHECK ALL ANSWERS!


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>