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Exact Solution of Cubic Equation

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					Chapter 03.02 Solution of Cubic Equations

After reading this chapter, you should be able to: 1. find the exact solution of a general cubic equation.

How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation ax 3  bx 2  cx  d  0 (1) 2 To find the roots of Equation (1), we first get rid of the quadratic term x  by making the substitution b (2) x y 3a to obtain

b  b  b     (3) a y    b y    c y    d  0 3a  3a  3a     Expanding Equation (3) and simplifying, we obtain the following equation   b2  2b 3 bc  ay 3   c   y   d   0 (4) 2   3a  3a  27a     Equation (4) is called the depressed cubic since the quadratic term is absent. Having the equation in this form makes it easier to solve for the roots of the cubic equation (Click here to know the history behind solving cubic equations exactly). First, convert the depressed cubic Equation (4) into the form 1 b2  1 2b 3 bc  3 y  c  y  d   0 2    a 3a  a 3a  27a 
y 3  ey  f  0

3

2

(5)

where

e

1 b2  c   a 3a   

03.02.1

03.02.2

Chapter 03.02

1 2b 3 bc  d    2  a 3a  27a  Now, reduce the above equation using Vieta’s substitution s yz z For the time being, the constant s is undefined. Substituting into the depressed cubic Equation (5), we get f 
3

(6)

s s   (7)  z    e z    f  0 z z   Expanding out and multiplying both sides by z 3 , we get z 6  3s  e z 4  fz 3  s3s  e z 2  s 3  0 (8) e Now, let s   ( s is no longer undefined) to simplify the equation into a tri-quadratic 3 equation. e3 z 6  fz 3  0 (9) 27 By making one more substitution, w  z 3 , we now have a general quadratic equation which can be solved using the quadratic formula. e3 w 2  fw  0 (10) 27 Once you obtain the solution to this quadratic equation, back substitute using the previous substitutions to obtain the roots to the general cubic equation. wz  y x where we assumed w  z3 (11) s yz z e (12) s 3 b x y 3a
Note: You will get two roots for w as Equation (10) is a quadratic equation. Using Equation (11) would then give you three roots for each of the two roots of w , hence giving you six root values for z . But the six root values of z would give you six values of y ( Equation (6) ); but three values of y will be identical to the other three. So one gets only three values of y , and hence three values of x . (Equation (2)) Example 1 Find the roots of the following cubic equation. x 3  9 x 2  36x  80  0

Solution of Cubic Equations Solution For the general form given by Equation (1) ax3  bx2  cx  d  0 we have a  1 , b  9 , c  36 , d  80 in x 3  9 x 2  36x  80  0 Equation (E1-1) is reduced to y 3  ey  f  0 where 1 b2  e  c   a 3a   

03.02.3

(E1-1)

 92 1   36  1 31  9
and

   

1 2b 3 bc  f  d    2  a 3a  27a  3 1 2 9  936      80   2 1 31  271    26
giving
y 3  9 y  26  0 For the general form given by Equation (5) y 3  ey  f  0 we have e  9 , f  26 in Equation (E1-2). From Equation (12) e s 3 9  3  3 From Equation (10) e3 w 2  fw  0 27 93 w 2  26 w  0 27 w 2  26w  27  0

(E1-2)

03.02.4 where
w  z3

Chapter 03.02

and
yz s z 3  z z

w

  26 

 262  41 27 21

 27,1 The solution is w1  27 w2  1 Since w  z3 z3  w For w  w1
z 3  w1  27  27e i 0

Since
w  z3

re i  uei  u 3e3i r cos  i sin    u 3 cos 3  i sin 3  resulting in r  u3 cos  cos3 sin   sin 3 Since sin  and cos are periodic of 2 , 3    2k   2k  3 k will take the value of 0, 1 and 2 before repeating the same values of  . So,   2k  , k  0, 1, 2 3  1  3   2  2  3
3





Solution of Cubic Equations

03.02.5

3 So roots of w  z 3 are

3 

  4 
1 3

   z1  r  cos  i sin  3 3  1   2   2   z 2  r 3  cos  i sin  3 3  

  4   4   z 3  r  cos  i sin  3 3  
1 3

gives
0 0 1/ 3  z1  27   cos  i sin  3 3  3 0  2 0  2  1/ 3  z 2  27   cos  i sin  3 3   2 2    3 cos  i sin  3 3  

 1 3   3   i  2 2   
3 3 3 i 2 2 0  4 0  4  1/ 3  z 3  27   cos  i sin  3 3   4 4    3 cos  i sin  3 3   

 1 3   3   i  2 2   
3 3 3   i 2 2

Since
yz 3 z

y1  z1 

3 z1 3  3 3 2

03.02.6
y2  z2  3 z2

Chapter 03.02

 3 3 3 3     i  2   3 3 3 2     i   2 2    5  3i 3  1  i 3

5  3i 3  1  i 3  1  i 3 1  i 3  1  i 2 3 3 y3  z3  z3 
 3 3 3 3     i  2   3 3 3 2     i   2 2    5  i3 3  1 i 3 5  i3 3 1  i 3   1 i 3 1 i 3  1  i 2 3
Since
x  y3 x1  y1  3  23 5 x2  y 2  3

  1  i2 3  3

 



 2  i2 3 x3  y 3  3  2  i2 3

  1  i2 3  3



The roots of the original cubic equation x 3  9 x 2  36x  80  0 are x1 , x2 , and x 3 , that is,

5 , 2  i2 3 , 2  i2 3 Verifying

Solution of Cubic Equations

03.02.7

x  5x  2  i 2
gives Using

3 x  2  i2 3  0

 



x 3  9 x 2  36x  80  0
w2  1 would yield the same values of the three roots of the equation. Try it.

Example 2 Find the roots of the following cubic equation x 3  0.03x 2  2.4  106  0 Solution For the general form ax 3  bx 2  cx  d  0 a  1, b  0.03, c  0, d  2.4  10 6 Depress the cubic equation by letting (Equation (2)) b x y 3a  0.03   y 31  y  0.01 Substituting the above equation into the cubic equation and simplifying, we get y 3  3  10 4 y  4  10 7   0 That gives e  3  10 4 and f  4 10 7 for Equation (5), that is, y 3  ey  f  0 . Now, solve the depressed cubic equation by using Vieta’s substitution as s yz z to obtain z 6  3s  3  10 4 z 4  4  10 7 z 3  s 3s  3  10 4 z 2  s 3  0 Letting e  3  10 4 s   10  4 3 3 we get the following tri-quadratic equation z 6  4  10 7 z 3  1 10 12  0 Using the following conversion, w  z 3 , we get a general quadratic equation w 2  4  10 7 w  1  10 12   0 Using the quadratic equation, the solutions for w are
w  4  10 7 

4 10 
21

7 2

 41 1  10 12





giving

03.02.8
w1  2  10 7  i 9.7979589711 3  10 7 w2  2  10
7

Chapter 03.02

  i 9.7979589711
3

3  10

7

 

Each solution of w  z yields three values of z . The three values of z from w1 are in rectangular form. Since w  z3 Then
zw
1 3

Let then

w  r cos  i sin    re i

z  u cos   i sin    ue i This gives w  z3

re i  uei  u 3e3i r cos  i sin    u 3 cos 3  i sin 3  resulting in r  u3 cos  cos3 sin   sin 3 Since sin  and cos are periodic of 2 , 3    2k   2k  3 k will take the value of 0, 1 and 2 before repeating the same values of  . So,   2k  , k  0, 1, 2 3  1  3   2  2  3   4  3  3 So the roots of w  z 3 are
3





   z1  r 3  cos  i sin  3 3  1   2   2   z 2  r 3  cos  i sin  3 3  
1

Solution of Cubic Equations

03.02.9

  4   4   z 3  r 3  cos  i sin  3 3  
1

So for
r
7 2

w1  2  10 7  i 9.7979589711 3  10 7

 2 10   9.79795897113 10 





7 2

 1 10 6 7 1 9.7979589711 3  10   tan  2  10 7  1.772154248 (2nd quadrant because y (the numerator) is positive and x (the denominator) is negative) 1 1.772154248 1.772154248   z1  1  10 6 3  cos  i sin  3 3    0.0083054095  i0.0055695756 17 35 1 1.772154248  2 1.772154248  2   z 2  1  10 6 3  cos  i sin  3 3    0.0089760987  i0.0044079078 46 15 1 1.772154248  4 1.772154248  4   z 3  1  10 6 3  cos  i sin  3 3    0.0006706892313  i0.009977483448 Compiling z1  0.0083054095 18  i0.0055695756 34 z 2  0.0089760987 46  i0.0044079078 14 z 3  6.7068922852 5  10 4  i 0.0099774834 48

Similarly, the three values of z from w2 in rectangular form are z 4  0.0083054095 18  i0.0055695756 34 z 5  0.0089760987 46  i 0.0044079078 14
z 6  6.7068922852 5  10 4  i 0.0099774834 48 Using Vieta’s substitution (Equation (6)), s yz z 110 4 y  z z we back substitute to find three values for y . For example, choosing z1  0.0083054095 18  i0.0055695756 34 gives





y1  0.0083054095 18  i 0.0055695756 34 

1  10 4 0.0083054095 18  i 0.0055695756 34

03.02.10
 0.0083054095 18  i 0.0055695756 34

Chapter 03.02

1  10  4 0.0083054095 18  i 0.0055695763 4  0.0083054090 78  i 0.0055695763 4 0.0083054095 18  i 0.0055695763 4  0.0083054095 18  i 0.0055695756 34 

1  10  4 0.0083054095 18  i0.0055695763 4 1  10  4  0.016610819036 The values of z1 , z 2 and z 3 give y1  0.0166108190 36 y 2  0.0179521974 9 y 3  0.0013413784 57 

respectively. The three other z values of z 4 , z 5 and z 6 give the same values as y1 , y 2 and y 3 , respectively. Now, using the substitution of x  y  0.01 the three roots of the given cubic equation are x1  0.0166108190 36  0.01  0.026610819036 x2  0.0179521974 9  0.01  0.00795219749 x3  0.0013413784 57  0.01  0.011341378457

NONLINEAR EQUATIONS Topic Exact Solution to Cubic Equations Summary Textbook notes on finding the exact solution to a cubic equation. Major General Engineering Authors Autar Kaw Last Revised December 17, 2009 Web Site http://numericalmethods.eng.usf.edu


				
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