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June 2nd, 2011, 12:45 PM   #1
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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
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June 2nd, 2011, 01:00 PM   #2
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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.
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June 2nd, 2011, 01:01 PM   #3
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Re: Solutions to the equation

x-3x+4x-2=0 and you already found out that x=1 fits, let's try to take that out of the equation.
x-3x+4x-2 | x-1
-(x-x).....|x - 2x + 2
-2x+4x-2
-(-2x+2x)
2x-2
-(2x-2)
0
So now we got something new: x-3x+4x-2 = (x-1)(x-2x+2)
Hope this is helping you out Now it's your task to find the solutions of this second degree polynomial
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June 2nd, 2011, 11:28 PM   #4
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Quote:
Originally Posted by Annelies 123
. . . x-3x+4x-2 = (x-1)(x-2x+2)
I'll assume that's correct. In that case, the discriminant for x - 2x + 2 equals to -4 < 0.
Use the quadratic formula to get x = (2 ?-4)/2 = (2 2i)/2 = 1 i.
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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.
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June 3rd, 2011, 12:17 AM   #6
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Re: Solutions to the equation

Great help guys

Thanks
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