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Solving Third-Degree Equations


Date: 09/24/2001 at 11:03:28
From: De Kort
Subject: Re: Solve third-degree equations

I'm trying to solve  x^3 + 2x^2 - 3x - 4 = 0
p = -13/3
q = -38/27
To calculate t I need to take the square root of  4p^3 + 27q^2, which
is -272... Of course I can consider t and u to be complex numbers, but 
in the end I'm looking for the  r e a l  solutions: and there are 
three of them!

How can I solve this?

Thanks,
Jan


Date: 09/25/2001 at 08:44:15
From: Doctor Jerry
Subject: Re: Solve third-degree equations

Hi Jan,

To refresh my memory of this I looked at Morris Kline's _Mathematical 
Thought from Ancient to Modern Times_, page 266.

Here's a sketch of Cardan's method - as completed by Euler. Start with 
the polynomial equation x^3 + a*x^2 + b*x + c = 0. Change variables by 
letting x=y - a/3. The polynomial equation becomes 

(1)                 y^3 + p*y + q = 0, 

where  p = b - a^2/3 and q = c - a*b/3 + 2a^3/27.  

Equation (1) can be solved by letting y = u + v, thus introducing, for 
a moment, the two new unknowns u and v. Upon substitution into the 
equation and rearranging the result, we find the equation
  
(2)             u^3 + v^3 + (p + 3*u*v)(u+v) + q = 0.

One condition on the two unknowns u and v is the requirement that  

(3)                p + 3*u*v = 0, 

which simplifies (2).  Equation (2) is now

(4)              u^3 + v^3 + q = 0.

Solving (3) for v in terms of u, substituting the result into (4), and 
multiplying by u^3 gives,

(5)                 (u^3)^2 + q(u^3) - p^3/27 = 0.                           

Solving (5) by the quadratic formula, we find the roots

               A = -q/2 + (q^2/4 + p^3/27)^{1/2} and 

               B = -q/2 - (q^2/4 + p^3/27)^{1/2},
               
where (q^2/4 + p^3/27)^{1/2} denotes either of the two square roots 
of q^2/4 + p^3/27.  Equation (5) is satisfied if we take u^3 = A and 
v^3 = B. Letting A^{1/3} denote a fixed (but arbitrary) cube root of 
A, then the three possible solutions of (5) are 

(5)   u_1 = A^{1/3},  u_2 =  w*A^{1/3},  u_3 = w^2*A^{1/3},

where w = (-1 + i*(1/2)3^{1/2}, one of the cube roots of 1. If we 
choose v_1 = B^{1/3} as that cube root of B for which  u_1*v_1 = -p/3, 
then corresponding to u_1, u_2, and u_3 will be 
v_1=B^{1/3}, v_2 = w^2*B^{1/3}, and v_3 = w*B^{1/3}. Equation (1) 
will then have the three roots  y_1 = u_1 + v_1, y_2 = u_2 + v_2, and 
y_3 = u_3 + v_3.   Finally, the roots to the equation 
x^3 + a*x^2 + b*x + c = 0 can be found easily.

The "irreducible case" is the case in which q^2/4 + p^3/27 < 0.  It 
can be shown that in the irreducible case all roots of (3) are real, 
although this is not obviously true from (5). There is a method due to 
Vieta, which simplifies the calculations in the irreducible case.  

I'll use the cubic x^3 + 2x^2 - 3x -4 = 0 you gave to illustrate. We 
first replace x by y - 2/3. This gives y^3 - (13/3)y - 38/27 = 0.  
Consider the trigonometric identity 

cos(3t) = 4*cos^3(t) - 3 cos(t) or, rearranging 

cos^3(t) - (3/4) cos(t) -(1/4)cos(3t) = 0.

Letting y = n*z, the equation y^3 - (13/3)y - 38/27 = 0 becomes

z^3 - (13/(3n^2)) z - 38/(27n^3) = 0.

We try to make this like the rearranged trig identity. Let 

- (13/(3n^2)) = -3/4 and -38/(27n^3) = -(1/4)cos(3t).

So,

n = 2*sqrt(13)/3  and  cos(3t) = 38/(26*sqrt(13)).

We find  t1 = 0.38447536423.  Calculating t2 = t1+2pi/3 and 
t3 = t1+4pi/3 we find roots

y1 = n*z1 = n*cos(t1) = 2.22821947948

y2 = n*z2 = n*cos(t2) = -1.8948864614

y3 = n*z3 = n*cos(t3) = -0.33333333333.

Of course, we could have noticed (using the rational roots theorem) 
that -1 was a root of the original equation. Using synthetic division 
to find the quadratic factor, we find the other roots to be
-1/2 +- sqrt(17)/2 .

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents
High School Polynomials

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