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 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^3-3x^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 x³-3x²+4x-2=0 and you already found out that x=1 fits, let's try to take that out of the equation. x³-3x²+4x-2 | x-1 -(x³-x²).....|x² - 2x + 2 -2x²+4x-2 -(-2x²+2x) 2x-2 -(2x-2) 0 So now we got something new: x³-3x²+4x-2 = (x-1)(x²-2x+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
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Quote:
 Originally Posted by Annelies 123 . . . x³-3x²+4x-2 = (x-1)(x²-2x+2)
I'll assume that's correct. In that case, the discriminant for x² - 2x + 2 equals to -4 < 0.
Use the quadratic formula to get x = (2 ± ?-4)/2 = (2 ± 2i)/2 = 1 ± i.

June 3rd, 2011, 12:07 AM   #5
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Re:

Quote:
 Originally Posted by johnny ... x = (2 ± ?-4)/2 = (2 ± 2i)/2 = 1 ± i.
I'll assume that's correct. In that case, we're done.

 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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