Quartic_Solver ( a, b, c , d, e, which_x )

Finds Real Roots of Quartic Equation a*x^4 + b*x^3 + c*x^2 + d*x + e = 0

Average rating: 3.5 (2 votes) Log in to vote

Dr. Ronald W. Satz - Show more from this author
Transpower Corporation
http://www.transpowercorp.com

Sample input:

Let
(
[
a = -20 ;
b = 5 ;
c = 17 ;
d = -29 ;
e = 87 ;
which_x = 1
] ;

Quartic_Solver ( a ; b ; c ; d ; e ; which_x )
)

Sample output:

1.48758311...

Function definition: (Copy & paste into FileMaker's Edit Custom Function window)

Let 
      (
        [
          delta_0 = SetPrecision ( ( c ^ 2 ) - 3 * b * d + 12 * a * e ; 30 ) ;
          delta_1 = SetPrecision ( 2 * ( c ^ 3 ) - 9 * b * c * d + 27 * ( b ^ 2 ) * e + 27 * a * ( d ^ 2 ) - 72 * a * c * e ; 30 ) ;
          // Q = (  .5 * ( delta_1 + Sqrt ( ( delta_1 ^ 2 ) - 4 * ( delta_0 ^ 3 )  ) ) ) ^ ( 1 / 3 ) ;
          Q_term1 = SetPrecision ( Sqrt ( ( delta_1 ^ 2 ) - 4 * ( delta_0 ^ 3 )  ) ; 30 ) ;
          Q_term2 = SetPrecision ( .5  * ( delta_1 + Q_term1 ) ; 30 ) ;
          Q = SetPrecision ( Q_term2 ^ ( 1 / 3 ) ; 30 ) ;
          p = SetPrecision ( ( 8 * a * c - 3 * ( b ^ 2 ) ) / ( 8 * ( a ^ 2 ) ) ; 30 ) ;
          qq = SetPrecision ( ( ( b ^ 3 ) - 4 * a * b * c + 8 * ( a ^ 2 ) * d ) / ( 8 * ( a ^ 3 ) ) ; 30 ) ;
         //  S = .5 * Sqrt ( ( - 2/3 ) * p + ( 1 / ( 3 * a ) )* ( Q + delta_0 / Q ) ) ;
          S_term1 = SetPrecision (  - ( 2 / 3 ) * p ; 30 ) ;
          S_term2 = SetPrecision ( 1 / ( 3 * a ) ; 30 ) ;
          S_term3 = SetPrecision ( Q + delta_0 / Q ; 30 ) ;
          S = SetPrecision ( .5 * Sqrt ( S_term1 + S_term2 * S_term3 ) ; 30 ) ;
          x1 = SetPrecision ( - ( b / ( 4 * a ) ) - S + .5 * Sqrt ( -4 * ( S ^ 2 ) - 2 * p + ( qq / S ) ) ; 30 ) ;
          x2 = SetPrecision (  - ( b / ( 4 * a ) ) - S  - .5 * Sqrt ( -4 * ( S ^ 2 ) - 2 * p + ( qq / S ) ) ; 30 ) ;
          x3 = SetPrecision ( - ( b / ( 4 * a ) )+ S + .5 * Sqrt ( -4 * ( S ^ 2 ) - 2 * p - ( qq / S ) ) ; 30 ) ;
          x4 = SetPrecision ( - ( b / ( 4 * a ) ) + S -  .5 * Sqrt ( -4 * ( S ^ 2 ) - 2 * p - ( qq / S ) ) ; 30 ) 
        ] ;
          
          Case
                   (
                      which_x = 1 ; x1 ;
                      which_x = 2 ; x2 ;
                      which_x = 3 ; x3 ;
                      which_x = 4 ; x4 
                   )
           )

You can, of course, change the precision to whatever you want (or even take it off altogether). If the x you want is complex, you'll simply get a "?" for the result. For most problems in physical science, you will want a real root. But you may have to run the program up to four times to get the root which makes physical sense. Ref.: Wikipedia ("Quartic Function")

Comments

Note: these functions are not guaranteed or supported by BrianDunning.com. Please contact the individual developer with any questions or problems.

Quartic_Solver ( a, b, c , d, e, which_x )

Comments

Support this website.

Support this website.