June 2nd, 2011, 12:45 PM  #1 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Solutions to the equation
Hey guys! I think i missed a lecture a while ago and i am struggling to recall how to find the solutions of eqautions, im sure it will come back to me, I have looked in my book but its not very clear. I have a past question, Find the solutions of the equation x^33x^2 +4x 2 =0 By inspection I can see x=1 is a solution, I recall having to compare coefficients? x^0 and x^1 but i forget why. Much appriciated. Jake 
June 2nd, 2011, 01:00 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond  Re: Solutions to the equation
One approach would be to divide x^3  3x^2 + 4x  2 by x  1. The result is a quadratic with no real roots, so x = 1 is the only real solution.

June 2nd, 2011, 01:01 PM  #3 
Newbie Joined: Jun 2011 Posts: 20 Thanks: 0  Re: Solutions to the equation
x³3x²+4x2=0 and you already found out that x=1 fits, let's try to take that out of the equation. x³3x²+4x2  x1 (x³x²).....x²  2x + 2 2x²+4x2 (2x²+2x) 2x2 (2x2) 0 So now we got something new: x³3x²+4x2 = (x1)(x²2x+2) Hope this is helping you out Now it's your task to find the solutions of this second degree polynomial 
June 2nd, 2011, 11:28 PM  #4  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
Use the quadratic formula to get x = (2 ± ?4)/2 = (2 ± 2i)/2 = 1 ± i.  
June 3rd, 2011, 12:07 AM  #5  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Quote:
 
June 3rd, 2011, 12:17 AM  #6 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Solutions to the equation
Great help guys Thanks 

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