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# Finding roots of a cubic equation

Nigelmca 16 June 2012 06:47 Posts5
Joined22/03/12
I am trying to follow through a worked example where real roots are expected, but I end up with imaginary roots. Where am I going wrong ? This is solving by a succession of formula substitutions rather than using the factor theorem.

so  a2=-2, a1=-5 and ao=6

Now substitute  where  z=(1/3)a2.  This results in equation of form x3+px=q where

p= -19/3  and q = - 56/27.

Now substitute x = w - p/3w to get equation of form w3 -  p3/(27w3) -  q  =  0.
Multiply through by w3 to get (w3)2 - q (w3) - (1/27)p3 = 0

I then get   (w3)2 +(56/27)w3 + (6859/729) = 0

This gives imaginary roots when applying the quadratic formula.

I have been through my workins a number of times but cannot find the error ! Help.

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stevencarrwork 16 June 2012 08:02 Posts680
Joined28/07/09
Isn't this why complex numbers were invented?

Because people were finding imaginary roots in steps in their working, when they knew that the final answer had to be a real number.
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Nigelmca 16 June 2012 08:45 Posts5
Joined22/03/12
OK Thanks Steven. I will continue working with imaginary numbers to try to get back to the real roots.
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davidboo 27 February 2018 06:23 Posts1
Joined27/02/18

Its very complex.

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