## Thursday, September 11, 2014

### Solutions to Cubic Equation Problems:

1) Check that for the roots obtained in the example problem:

a)     (a1 x a2) + (a 2 x a3) + (a3 x a1) = q

AND

b) a1 x a2 x a3 = r

Solution:   From the example problem we saw:

a1 = 1.103 - 0.665i

a2 = 1.103 + 0.665i

a3 = -1.206

Then: (a1 x a2) = (1.103 - 0.665i) (1.103 + 0.665i ) = 1.659

(a 2 x a3)  =   (1.103 + 0.665i) (-1.206)  =  -1.33 – 0.802i

(a3 x a1) =   (-1.206) (1.103 - 0.665i =  -1.33 + 0.802i

So:  (a1 x a2) + (a 2 x a3) + (a3 x a1) =

1.659 + (-1.33 – 0.802i) + (-1.33 + 0.802i) »  1.66 – 2.66 =  -1 = q

b) a1 x a2 x a3 = (1.103 - 0.665i) (1.103 + 0.665i)( -1.206) » - 2 = r

2) Solve: x3 – 6x2 + 11x – 6

Then: p = 6, q = -11, r = 6

D =  [(- 4p3 r – 27r2 + 18 pqr – 4 q3 + p2 q2)] 1/2

r 1=   - 1/2 + 1/2  Ö- 3

r 2=   - 1/2 -  1/2  Ö- 3

So:  D = 60.033i

x1 = [p3 –9/2 (pq – 3r) – 3/2 x (-3)1/2 x D ]1/3

=  9.085

x2 = (p2 – 3q)/ x1

=   7.595

The roots are then:

a1 = 1/3(p + r2 x x1 + r x x2)   =   7.56

a2 = 1/3( p + r2 x x1+ r x x2)   =  -0.78 – 0.43i

a3 = 1/3( p + r x x1 + rx r2 x x2)

=   -0.78 + 0.43i

Check: a1 + a2 + a3 = p =  6

=  7.56 +  (-0.78 – 0.43i)  + ( -0.78 + 0.43i) =  7.56 - 1.56 = 6

(a1 x a2) + (a 2 x a3) + (a3 x a1) = q = -11

(a1 – a2) x (a2 – a3) x (a3 – a1) = D = 60.033i

Q.E.D.