Roots of a quadratic function

Roots of a quadratic function
You are encouraged to solve this task according to the task description, using any language you may know.

This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.

Write a program to find the roots of a quadratic equation, i.e., solve the equation $ax^{2}+bx+c=0$. Your program must correctly handle non-real roots, but it need not check that $a\neq 0$.

The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic. The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other. In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with $a=1$, $b=-10^{5}$, and $c=1$. (For double-precision floats, set $b=-10^{9}$.) Consider the following implementation in Ada: <lang ada>with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;

procedure Quadratic_Equation is

  type Roots is array (1..2) of Float;
  function Solve (A, B, C : Float) return Roots is
     SD : constant Float := sqrt (B**2 - 4.0 * A * C);
     AA : constant Float := 2.0 * A;
  begin
     return ((- B + SD) / AA, (- B - SD) / AA);
  end Solve;

  R : constant Roots := Solve (1.0, -10.0E5, 1.0);

begin

  Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));

end Quadratic_Equation;</lang> Sample output:

X1 = 1.00000E+06 X2 = 0.00000E+00

As we can see, the second root has lost all significant figures. The right answer is that X2 is about $10^{-6}$. The naive method is numerically unstable.

Task: do it better. This means that given $a=1$, $b=-10^{9}$, and $c=1$, both of the roots your program returns should be greater than $10^{-11}$. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle $b=-10^{6}$. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of "What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.

Ada

<lang ada>with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;

procedure Quadratic_Equation is

  type Roots is array (1..2) of Float;
  function Solve (A, B, C : Float) return Roots is
     SD : constant Float := sqrt (B**2 - 4.0 * A * C);
     X  : Float;
  begin
     if B < 0.0 then
        X := (- B + SD) / 2.0 * A;
        return (X, C / (A * X));
     else
        X := (- B - SD) / 2.0 * A;
        return (C / (A * X), X);
     end if;
  end Solve;

  R : constant Roots := Solve (1.0, -10.0E5, 1.0);

begin

  Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2))); 

end Quadratic_Equation;</lang> Here precision loss is prevented by checking signs of operands. On errors, Constraint_Error is propagated on numeric errors and when roots are complex. Sample output:

X1 = 1.00000E+06 X2 = 1.00000E-06

ALGOL 68

Translation of: Ada

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

<lang algol68>quadratic equation: BEGIN

 MODE ROOTS  = UNION([]REAL, []COMPL);
 MODE QUADRATIC = STRUCT(REAL a,b,c);

 PROC solve  = (QUADRATIC q)ROOTS:
 BEGIN
   REAL a = a OF q, b = b OF q, c = c OF q;
   REAL sa = b**2 - 4*a*c;
   IF sa >=0 THEN # handle the +ve case as REAL #
     REAL sqrt sa = ( b<0 | sqrt(sa) | -sqrt(sa));
     REAL r1 = (-b + sqrt sa)/(2*a),
          r2 = (-b - sqrt sa)/(2*a);
     []REAL((r1,r2))
   ELSE # handle the -ve case as COMPL conjugate pairs #
     COMPL compl sqrt sa = ( b<0 | complex sqrt(sa) | -complex sqrt(sa));
     COMPL r1 = (-b + compl sqrt sa)/(2*a),
           r2 = (-b - compl sqrt sa)/(2*a);
     []COMPL (r1, r2)
   FI
 END # solve #;

 PROC real  evaluate = (QUADRATIC q, REAL x )REAL:  (a OF q*x + b OF q)*x + c OF q;
 PROC compl evaluate = (QUADRATIC q, COMPL x)COMPL: (a OF q*x + b OF q)*x + c OF q;

 # only a very tiny difference between the 2 examples #
 []QUADRATIC test = ((1, -10e5, 1), (1, 0, 1), (1,-3,2), (1,3,2), (4,0,4), (3,4,5));

 FORMAT real fmt = $g(-0,8)$;
 FORMAT compl fmt = $f(real fmt)"+"f(real fmt)"i"$;
 FORMAT quadratic fmt = $f(real fmt)" x**2 + "f(real fmt)" x + "f(real fmt)" = 0"$;

 FOR index TO UPB test DO
   QUADRATIC quadratic = test[index];
   ROOTS r = solve(quadratic);

1. Output the two different scenerios #

   printf(($"Quadratic: "$, quadratic fmt, quadratic, $l$));
   CASE r IN
     ([]REAL r): 
       printf(($"REAL x1 = "$, real fmt, r[1],
                  $", x2 = "$, real fmt, r[2],  $"; "$,
               $"REAL y1 = "$, real fmt, real evaluate(quadratic,r[1]),
                  $", y2 = "$, real fmt, real evaluate(quadratic,r[2]), $";"ll$
       )),
     ([]COMPL c):
       printf(($"COMPL x1,x2 = "$, real fmt, re OF c[1], $"+/-"$, 
                                   real fmt, ABS im OF c[1], $"; "$,
                 $"COMPL y1 = "$, compl fmt, compl evaluate(quadratic,c[1]),
                     $", y2 = "$, compl fmt, compl evaluate(quadratic,c[2]), $";"ll$
       ))
   ESAC
 OD

END # quadratic_equation #</lang> Output:

Quadratic: 1.00000000 x**2 + -1000000.00000000 x + 1.00000000 = 0
REAL x1 = 999999.99999900, x2 = .00000100; REAL y1 = -.00000761, y2 = -.00000761;

Quadratic: 1.00000000 x**2 + .00000000 x + 1.00000000 = 0
COMPL x1,x2 = .00000000+/-1.00000000; COMPL y1 = .00000000+.00000000i, y2 = .00000000+.00000000i;

Quadratic: 1.00000000 x**2 + -3.00000000 x + 2.00000000 = 0
REAL x1 = 2.00000000, x2 = 1.00000000; REAL y1 = .00000000, y2 = .00000000;

Quadratic: 1.00000000 x**2 + 3.00000000 x + 2.00000000 = 0
REAL x1 = -2.00000000, x2 = -1.00000000; REAL y1 = .00000000, y2 = .00000000;

Quadratic: 4.00000000 x**2 + .00000000 x + 4.00000000 = 0
COMPL x1,x2 = .00000000+/-1.00000000; COMPL y1 = .00000000+.00000000i, y2 = .00000000+.00000000i;

Quadratic: 3.00000000 x**2 + 4.00000000 x + 5.00000000 = 0
COMPL x1,x2 = -.66666667+/-1.10554160; COMPL y1 = .00000000+.00000000i, y2 = .00000000+-.00000000i;

AutoHotkey

ahk forum: discussion <lang AutoHotkey>MsgBox % quadratic(u,v, 1,-3,2) ", " u ", " v MsgBox % quadratic(u,v, 1,3,2) ", " u ", " v MsgBox % quadratic(u,v, -2,4,-2) ", " u ", " v MsgBox % quadratic(u,v, 1,0,1) ", " u ", " v SetFormat FloatFast, 0.15e MsgBox % quadratic(u,v, 1,-1.0e8,1) ", " u ", " v

quadratic(ByRef x1, ByRef x2, a,b,c) { ; -> #real roots {x1,x2} of ax²+bx+c

  If (a = 0)
     Return -1 ; ERROR: not quadratic
  d := b*b - 4*a*c
  If (d < 0) {
     x1 := x2 := ""
     Return 0
  }
  If (d = 0) {
     x1 := x2 := -b/2/a
     Return 1
  }
  x1 := (-b - (b<0 ? -sqrt(d) : sqrt(d)))/2/a
  x2 := c/a/x1
  Return 2

}</lang>

C

<lang c>#include <stdio.h>

1. include <math.h>
2. include <complex.h>

void roots_quadratic_eq(double a, double b, double c, complex double *x) {

 double delta;

 delta = b*b - 4.0*a*c;
 x[0] = (-b + csqrt(delta)) / (2.0*a);
 x[1] = (-b - csqrt(delta)) / (2.0*a);

}</lang>

Translation of: C++

<lang c>void roots_quadratic_eq2(double a, double b, double c, complex double *x) {

 b /= a;
 c /= a;
 double delta = b*b - 4*c;
 if ( delta < 0 ) {
   x[0] = -b/2 + I*sqrt(-delta)/2.0;
   x[1] = -b/2 - I*sqrt(-delta)/2.0;
 } else {
   double root = sqrt(delta);
   double sol = (b>0) ? (-b - root)/2.0 : (-b + root)/2.0;
   x[0] = sol;
   x[1] = c/sol;
 }

}</lang>

<lang c>int main() {

 complex double x[2];

 roots_quadratic_eq(1, -1e20, 1, x);
 printf("x1 = (%.20le, %.20le)\nx2 = (%.20le, %.20le)\n\n",

creal(x[0]), cimag(x[0]), creal(x[1]), cimag(x[1]));

 roots_quadratic_eq2(1, -1e20, 1, x);
 printf("x1 = (%.20le, %.20le)\nx2 = (%.20le, %.20le)\n\n",

creal(x[0]), cimag(x[0]), creal(x[1]), cimag(x[1]));

 return 0;

}</lang>

x1 = (1.00000000000000000000e+20, 0.00000000000000000000e+00)
x2 = (0.00000000000000000000e+00, 0.00000000000000000000e+00)

x1 = (1.00000000000000000000e+20, 0.00000000000000000000e+00)
x2 = (9.99999999999999945153e-21, 0.00000000000000000000e+00)

C#

<lang csharp>using System; using System.Numerics;

class QuadraticRoots {

   static Tuple<Complex, Complex> Solve(double a, double b, double c)
   {
       var q = -(b + Math.Sign(b) * Complex.Sqrt(b * b - 4 * a * c)) / 2;
       return Tuple.Create(q / a, c / q);
   }

   static void Main()
   {
       Console.WriteLine(Solve(1, -1E20, 1));
   }

}</lang> Output: <lang>((1E+20, 0), (1E-20, 0))</lang>

C++

<lang cpp>#include <iostream>

1. include <utility>
2. include <complex>

typedef std::complex<double> complex;

std::pair<complex, complex>

solve_quadratic_equation(double a, double b, double c)

{

 b /= a;
 c /= a;
 double discriminant = b*b-4*c;
 if (discriminant < 0)
   return std::make_pair(complex(-b/2, std::sqrt(-discriminant)/2),
                         complex(-b/2, -std::sqrt(-discriminant)/2));

 double root = std::sqrt(discriminant);
 double solution1 = (b > 0)? (-b - root)/2
                           : (-b + root)/2;

 return std::make_pair(solution1, c/solution1);

}

int main() {

 std::pair<complex, complex> result = solve_quadratic_equation(1, -1e20, 1);
 std::cout << result.first << ", " << result.second << std::endl;

}</lang> Output:

(1e+20,0), (1e-20,0)

Common Lisp

<lang lisp>(defun quadratic (a b c)

 "Compute the roots of a quadratic in the form ax^2 + bx + c = 1.  Evaluates to a list of the two roots."
 (let ((discriminant (- (expt b 2) (* 4 a c)))
       (denominator (* 2 a))
       (neg-b (* b -1)))
   (list (/ (+ neg-b (sqrt discriminant)) denominator)
    (/ (- neg-b (sqrt discriminant)) denominator))))</lang>

D

<lang d>import std.stdio, std.math, std.traits ;

auto naiveQR(T,U,V)(T a, U b, V c) if(isFloatingPoint!T) {

   alias CommonType!(T,U,V) RT ;
   if(a == 0) return [cast(RT)c/b] ; // linear
   RT det = b*b - 4 * a * c ;
   if(det < 0) return cast(RT[])[] ; // no real number root
   immutable SD = sqrt(det) ;

   return [(-b + SD) / 2 * a, (-b - SD) / 2 * a] ;

}

auto cautiQR(T,U,V)(T a, U b, V c) if(isFloatingPoint!T) {

   alias CommonType!(T,U,V) RT ;
   if(a == 0) return [cast(RT)c/b] ; // linear
   RT det = b*b - 4 * a * c ;
   if(det < 0) return cast(RT[])[] ; // no real number root
   immutable SD = sqrt(det) ;

   if(b*a < 0) {
       auto x = (-b + SD) / 2 * a ;
       return [x, c/(a * x)] ;
   } else {
       auto x = (-b - SD) / 2 * a ;
       return [c/(a * x), x] ;
   }

}

void main() {

   writefln("   naive : %6g", naiveQR(1F,-10e5F,1)) ; // force use 32bit float
   writefln("cautious : %6g", cautiQR(1F,-10e5F,1)) ;

}</lang> output:

   naive : [ 1e+06,      0]
cautious : [ 1e+06,  1e-06]

Factor

Translation of: Ada

<lang factor>:: quadratic-equation ( a b c -- x1 x2 )

   b sq a c * 4 * - sqrt :> sd
   b 0 <
   [ b neg sd + a 2 * / ]
   [ b neg sd - a 2 * / ] if :> x
   x c a x * / ;</lang>

<lang factor>( scratchpad ) 1 -1.e20 1 quadratic-equation --- Data stack: 1.0e+20 9.999999999999999e-21</lang>

Forth

Without locals: <lang forth>: quadratic ( fa fb fc -- r1 r2 )

 frot frot
 ( c a b )
 fover 3 fpick f* -4e f*  fover fdup f* f+
 ( c a b det )
 fdup f0< if abort" imaginary roots" then
 fsqrt
 fover f0< if fnegate then
 f+ fnegate
 ( c a b-det )
 2e f/ fover f/  
 ( c a r1 )
 frot frot f/ fover f/ ;</lang>

With locals: <lang forth>: quadratic { F: a F: b F: c -- r1 r2 }

 b b f*  4e a f* c f* f-
 fdup f0< if abort" imaginary roots" then
 fsqrt
 b f0< if fnegate then b f+ fnegate 2e f/ a f/
 c a f/ fover f/ ;

\ test 1 set-precision 1e -1e6 1e quadratic fs. fs. \ 1e-6 1e6</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>PROGRAM QUADRATIC

IMPLICIT NONE
INTEGER, PARAMETER :: dp = SELECTED_REAL_KIND(15)
REAL(dp) :: a, b, c, e, discriminant, rroot1, rroot2
COMPLEX(dp) :: croot1, croot2

WRITE(*,*) "Enter the coefficients of the equation ax^2 + bx + c"
WRITE(*, "(A)", ADVANCE="NO") "a = "
READ *, a
WRITE(*,"(A)", ADVANCE="NO") "b = "
READ *, b
WRITE(*,"(A)", ADVANCE="NO") "c = "
READ *, c

WRITE(*,"(3(A,E23.15))") "Coefficients are: a = ", a, "   b = ", b, "   c = ", c
e = 1.0e-9_dp
discriminant = b*b - 4.0_dp*a*c

IF (ABS(discriminant) < e) THEN
   rroot1 = -b / (2.0_dp*a)
   WRITE(*,*) "The roots are real and equal:"
   WRITE(*,"(A,E23.15)") "Root = ", rroot1
ELSE IF (discriminant > 0) THEN
   IF (b < 0) THEN
      rroot1 = (-b + SQRT(discriminant)) / (2.0_dp*a)
      rroot2 = c / (a * rroot1)
   ELSE
      rroot1 = (-b - SQRT(discriminant)) / (2.0_dp*a)
      rroot2 = c / (a * rroot1)
   END IF
   WRITE(*,*) "The roots are real:"
   WRITE(*,"(2(A,E23.15))") "Root1 = ", rroot1, "  Root2 = ", rroot2
ELSE
   croot1 = (-b + SQRT(CMPLX(discriminant))) / (2.0_dp*a) 
   croot2 = CONJG(croot1)
   WRITE(*,*) "The roots are complex:" 
   WRITE(*,"(2(A,2E23.15,A))") "Root1 = ", croot1, "j ", "  Root2 = ", croot2, "j"
END IF</lang>

Sample output

Coefficients are: a =   0.300000000000000E+01   b =   0.400000000000000E+01   c =   0.133333333330000E+01
The roots are real and equal:
Root =  -0.666666666666667E+00

Coefficients are: a =   0.300000000000000E+01   b =   0.200000000000000E+01   c =  -0.100000000000000E+01
The roots are real:
Root1 =  -0.100000000000000E+01  Root2 =   0.333333333333333E+00

Coefficients are: a =   0.300000000000000E+01   b =   0.200000000000000E+01   c =   0.100000000000000E+01
The roots are complex:
Root1 =  -0.333333333333333E+00  0.471404512723287E+00j   Root2 =  -0.333333333333333E+00 -0.471404512723287E+00j

Coefficients are: a =   0.100000000000000E+01   b =  -0.100000000000000E+07   c =   0.100000000000000E+01
The roots are real:
Root1 =   0.999999999999000E+06  Root2 =   0.100000000000100E-05

GAP

<lang gap>QuadraticRoots := function(a, b, c)

 local d;
 d := Sqrt(b*b - 4*a*c);
 return [ (-b+d)/(2*a), (-b-d)/(2*a) ];

end;

1. Hint : E(12) is a 12th primitive root of 1

QuadraticRoots(2, 2, -1);

1. [ 1/2*E(12)^4-1/2*E(12)^7+1/2*E(12)^8+1/2*E(12)^11,
2. 1/2*E(12)^4+1/2*E(12)^7+1/2*E(12)^8-1/2*E(12)^11 ]</lang>

Go

<lang go>package main

import (

   "fmt"
   "math"

)

func qr(a, b, c float64) ([]float64, []complex128) {

   d := b*b-4*a*c
   switch {
   case d == 0:
       // single root
       return []float64{-b/(2*a)}, nil
   case d > 0:
       // two real roots
       if b < 0 {
           d = math.Sqrt(d)-b
       } else {
           d = -math.Sqrt(d)-b
       }
       return []float64{d/(2*a), (2*c)/d}, nil
   case d < 0:
       // two complex roots

       den := 1/(2*a)
       t1 := complex(-b*den, 0)
       t2 := complex(0, math.Sqrt(-d)*den)
       return nil, []complex128{t1+t2, t1-t2}
   }
   // otherwise d overflowed or a coefficient was NAN
   return []float64{d}, nil

}

func test(a, b, c float64) {

   fmt.Print("coefficients: ", a, b, c, " -> ")
   r, i := qr(a, b, c)
   switch len(r) {
   case 1:
       fmt.Println("one real root:", r[0])
   case 2:
       fmt.Println("two real roots:", r[0], r[1])
   default:
       fmt.Println("two complex roots:", i[0], i[1])
   }

}

func main() {

   var zero float64
   for _, c := range [][3]float64{
       {1, -2, 1},
       {1, 0, 1},
       {1, -10, 1},
       {1, -1000, 1},
       {1, -1e9, 1},
   } {
       test(c[0], c[1], c[2])
   }

}</lang> Output:

coefficients: 1 -2 1 -> one real root: 1
coefficients: 1 0 1 -> two complex roots: (0+1i) (-0-1i)
coefficients: 1 -10 1 -> two real roots: 9.898979485566356 0.10102051443364381
coefficients: 1 -1000 1 -> two real roots: 999.998999999 0.001000001000002
coefficients: 1 -1e+09 1 -> two real roots: 1e+09 1e-09

Haskell

<lang haskell>import Data.Complex

type CD = Complex Double

quadraticRoots :: (CD, CD, CD) -> (CD, CD) quadraticRoots (a, b, c) =

   if   realPart b > 0
   then ((2*c) / (-b - d), (-b - d) / (2*a))
   else ((-b + d) / (2*a), (2*c) / (-b + d))
 where d = sqrt $ b^2 - 4*a*c</lang>

*Main> mapM_ print $ map quadraticRoots [(3, 4, 4/3), (3, 2, -1), (3, 2, 1), (1, -10e5, 1), (1, -10e9, 1)]
((-0.6666666666666666) :+ (-0.0),(-0.6666666666666666) :+ 0.0)
(0.3333333333333333 :+ 0.0,(-1.0) :+ 0.0)
((-0.33333333333333326) :+ 0.4714045207910316,(-0.3333333333333333) :+ (-0.47140452079103173))
(999999.999999 :+ 0.0,1.000000000001e-6 :+ 0.0)
(1.0e10 :+ 0.0,1.0e-10 :+ 0.0)

IDL

<lang idl>compile_OPT IDL2

print, "input a, press enter, input b, press enter, input c, press enter" read,a,b,c Promt='Enter values of a,b,c and hit enter'

a0=0.0 b0=0.0 c0=0.0 ;make them floating point variables

x=-b+sqrt((b^2)-4*a*c) y=-b-sqrt((b^2)-4*a*c) z=2*a d= x/z e= y/z

print, d,e</lang>

J

Solution use J's built-in polynomial solver:

   p.

Example using inputs from other solutions and the unstable example from the task description:

<lang j> coeff =. _3 |.\ 3 4 4r3 3 2 _1 3 2 1 1 _1e6 1

  > {:"1 p. coeff
        _0.666667           _0.666667
               _1            0.333333

_0.333333j0.471405 _0.333333j_0.471405

              1e6                1e_6</lang>

Of course p. generalizes to polynomials of any order. Given the coefficients p. returns the multiplier and roots of the polynomial. Given the multiplier and roots it returns the coefficients. For example using the cubic $0+16x-12x^{2}+2x^{3}$: <lang j> p. 0 16 _12 2 NB. return multiplier ; roots +-+-----+ |2|4 2 0| +-+-----+

  p. 2 ; 4 2 0    NB. return coefficients

0 16 _12 2</lang>

Exploring the limits of precision:

<lang j> 1{::p. 1 _1e5 1 NB. display roots 100000 1e_5

  1 _1e5 1 p. 1{::p. 1 _1e5 1  NB. test roots

_3.38436e_7 0

  1 _1e5 1 p. 1e5 1e_5  NB. test displayed roots

1 9.99999e_11

  1e5 1e_5 - 1{::p. 1 _1e5 1  NB. find difference

1e_5 _1e_15

  1 _1e5 1 p. 1e5 1e_5-1e_5 _1e_15  NB. test displayed roots with adjustment

_3.38436e_7 0</lang>

Java

<lang java>public class Main {

   static class Complex {
       double re, im;

       public Complex(double re, double im) {
           this.re = re;
           this.im = im;
       }

       @Override
       public boolean equals(Object obj) {
           if (obj == this) {return true;}
           if (!(obj instanceof Complex)) {return false;}
           Complex other = (Complex) obj;
           return (re == other.re) && (im == other.im);
       }

       @Override
       public String toString() {
           if (im == 0.0) {return String.format("%g", re);}
           if (re == 0.0) {return String.format("%gi", im);}
           return String.format("%g %c %gi", re,
               (im < 0.0 ? '-' : '+'), Math.abs(im));
       }
   }

   private static Complex[] quadraticRoots(double a, double b, double c) {
       Complex[] roots = new Complex[2];
       double d = b * b - 4.0 * a * c;
       double aa = a + a;
       if (d < 0.0) {
           double re = -b / aa;
           double im = Math.sqrt(-d) / aa;
           roots[0] = new Complex(re, im);
           roots[1] = new Complex(re, -im);
       } else {
           roots[0] = new Complex((-b + Math.sqrt(d)) / aa, 0.0);
           roots[1] = new Complex((-b - Math.sqrt(d)) / aa, 0.0);
       }
       return roots;
   }

   public static void main(String[] args) {
       double[][] equations = {
           {1.0, -10.0e5, 1.0},    // two distinct real roots
           {1.0, 2.0, 1.0},        // one real root (double root)
           {1.0, 0.0, 1.0},        // two imaginary roots
           {1.0, 1.0, 1.0}         // two complex roots
       };
       for (int i = 0; i < equations.length; i++) {
           Complex[] roots = quadraticRoots(
               equations[i][0], equations[i][1], equations[i][2]);
           System.out.format("%na = %g   b = %g   c = %g%n",
               equations[i][0], equations[i][1], equations[i][2]);
           if (roots[0].equals(roots[1])) {
               System.out.format("X1,2 = %s%n", roots[0]);
           } else {
               System.out.format("X1 = %s%n", roots[0]);
               System.out.format("X2 = %s%n", roots[1]);
           }
       }
   }

}</lang> Output:

a = 1.00000   b = -1.00000e+06   c = 1.00000
X1 = 1.00000e+06
X2 = 1.00001e-06

a = 1.00000   b = 2.00000   c = 1.00000
X1,2 = -1.00000

a = 1.00000   b = 0.00000   c = 1.00000
X1 = 1.00000i
X2 = -1.00000i

a = 1.00000   b = 1.00000   c = 1.00000
X1 = -0.500000 + 0.866025i
X2 = -0.500000 - 0.866025i

Liberty BASIC

<lang lb>a=1:b=2:c=3

   'assume a<>0
   print quad$(a,b,c)
   end

function quad$(a,b,c)

   D=b^2-4*a*c
   x=-1*b
   if D<0 then
       quad$=str$(x/(2*a));" +i";str$(sqr(abs(D))/(2*a));" , ";str$(x/(2*a));" -i";str$(sqr(abs(D))/abs(2*a))
   else
       quad$=str$(x/(2*a)+sqr(D)/(2*a));" , ";str$(x/(2*a)-sqr(D)/(2*a))
   end if

end function</lang>

Logo

<lang logo>to quadratic :a :b :c

 localmake "d sqrt (:b*:b - 4*:a*:c)
 if :b < 0 [make "d minus :d]
 output list (:d-:b)/(2*:a) (2*:c)/(:d-:b)

end</lang>

Lua

In order to correctly handle complex roots, qsolve must be given objects from a suitable complex number library, like that from the Complex Numbers article. However, this should be enough to demonstrate its accuracy:

<lang lua>function qsolve(a, b, c)

 if b < 0 then return qsolve(-a, -b, -c) end
 val = b + (b^2 - 4*a*c)^(1/2) --this never exhibits instability if b > 0
 return -val / (2 * a), -2 * c / val --2c / val is the same as the "unstable" second root

end

for i = 1, 12 do

 print(qsolve(1, 0-10^i, 1))

end</lang> The "trick" lies in avoiding subtracting large values that differ by a small amount, which is the source of instability in the "normal" formula. It is trivial to prove that 2c/(b + sqrt(b^2-4ac)) = (b - sqrt(b^2-4ac))/2a.

Mathematica

Possible ways to do this are (symbolic and numeric examples): <lang Mathematica>Solve[a x^2 + b x + c == 0, x] Solve[x^2 - 10^5 x + 1 == 0, x] Root[#1^2 - 10^5 #1 + 1 &, 1] Root[#1^2 - 10^5 #1 + 1 &, 2] Reduce[a x^2 + b x + c == 0, x] Reduce[x^2 - 10^5 x + 1 == 0, x] FindInstance[x^2 - 10^5 x + 1 == 0, x, Reals, 2] FindRoot[x^2 - 10^5 x + 1 == 0, {x, 0}] FindRoot[x^2 - 10^5 x + 1 == 0, {x, 10^6}]</lang> gives back:

Some lines in this example are too long (more than 80 characters). Please fix the code if it's possible and remove this message.

$\left\{\left\{x\to {\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\right\},\left\{x\to {\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\right\}\right\}$

$\left\{\left\{x\to {\frac {1}{50000+{\sqrt {2499999999}}}}\right\},\left\{x\to 50000+{\sqrt {2499999999}}\right\}\right\}$

$50000-{\sqrt {2499999999}}$

$50000+{\sqrt {2499999999}}$

${\begin{aligned}{\Biggl (}a&\neq 0\And \And \left(x=={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\|x=={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\right){\Biggr )}\\&{\biggl \|}\left(a==0\And \And b\neq 0\And \And x==-{\frac {c}{b}}\right)\\&{\biggr \|}(c==0\And \And b==0\And \And a==0)\end{aligned}}$

$x=={\frac {1}{50000+{\sqrt {2499999999}}}}\|x==50000+{\sqrt {2499999999}}$

$\left\{\left\{x\to {\frac {1}{50000+{\sqrt {2499999999}}}}\right\},\left\{x\to 50000+{\sqrt {2499999999}}\right\}\right\}$

$\{x\to 0.00001\}$

$\{x\to 100000.\}$

Note that some functions do not really give the answer (like reduce) rather it gives another way of writing it (boolean expression). However note that reduce gives the explicit cases for a zero and nonzero, b zero and nonzero, et cetera. Some functions are numeric by nature, other can handle both symbolic and numeric. In generals the solution will be exact if the input is exact. Any exact result can be approximated to arbitrary precision using the function N[expression,number of digits]. Further notice that some functions give back exact answers in a different form then others, however the answers are both correct, the answers are just written differently.

MATLAB

<lang Matlab>roots([1 -3 2])  % coefficients in decreasing order of power e.g. [x^n ... x^2 x^1 x^0]</lang>

Modula-3

Translation of: Ada

<lang modula3>MODULE Quad EXPORTS Main;

IMPORT IO, Fmt, Math;

TYPE Roots = ARRAY [1..2] OF LONGREAL;

VAR r: Roots;

PROCEDURE Solve(a, b, c: LONGREAL): Roots =

 VAR sd: LONGREAL := Math.sqrt(b * b - 4.0D0 * a * c);
     x: LONGREAL;
 BEGIN
   IF b < 0.0D0 THEN
     x := (-b + sd) / 2.0D0 * a;
     RETURN Roots{x, c / (a * x)};
   ELSE
     x := (-b - sd) / 2.0D0 * a;
     RETURN Roots{c / (a * x), x};
   END;
 END Solve;

BEGIN

 r := Solve(1.0D0, -10.0D5, 1.0D0);
 IO.Put("X1 = " & Fmt.LongReal(r[1]) & " X2 = " & Fmt.LongReal(r[2]) & "\n");

END Quad.</lang>

OCaml

<lang ocaml>type quadroots =

 | RealRoots of float * float
 | ComplexRoots of Complex.t * Complex.t ;;

let quadsolve a b c =

 let d = (b *. b) -. (4.0 *. a *. c) in
 if d < 0.0
 then
   let r = -. b /. (2.0 *. a)
   and i = sqrt(-. d) /. (2.0 *. a) in
   ComplexRoots ({ Complex.re = r; Complex.im = i },
                 { Complex.re = r; Complex.im = (-.i) })
 else
   let r = if b < 0.0
   then ((sqrt d) -. b) /. (2.0 *. a)
   else ((sqrt d) +. b) /. (-2.0 *. a)
 in
 RealRoots (r, c /. (r *. a))

</lang>

Sample output: <lang ocaml># quadsolve 1.0 0.0 (-2.0) ;; - : quadroots = RealRoots (-1.4142135623730951, 1.4142135623730949)

1. quadsolve 1.0 0.0 2.0 ;;

- : quadroots = ComplexRoots ({Complex.re = 0.; Complex.im = 1.4142135623730951},

{Complex.re = 0.; Complex.im = -1.4142135623730951})

1. quadsolve 1.0 (-1.0e5) 1.0 ;;

- : quadroots = RealRoots (99999.99999, 1.0000000001000001e-005)</lang>

Octave

See MATLAB.

PARI/GP

Translation of: C

<lang>roots(a,b,c)={

 b /= a;
 c /= a;
 my (delta = b^2 - 4*c, root=sqrt(delta));
 if (delta < 0,
   [root-b,-root-b]/2
 ,
   my(sol=if(b>0, -b - root,-b + root)/2);
   [sol,c/sol]
 )

};</lang>

Perl

When using Math::Complex perl automatically convert numbers when necessary. <lang perl>use Math::Complex;

($x1,$x2) = solveQuad(1,2,3);

print "x1 = $x1, x2 = $x2\n";

sub solveQuad { my ($a,$b,$c) = @_; my $root = sqrt($b**2 - 4*$a*$c); return ( -$b + $root )/(2*$a), ( -$b - $root )/(2*$a); }</lang>

Perl 6

Perl 6 has complex number handling built in.

<lang perl6>my @sets = [1, 2, 1],

          [1, 2, 3],
          [1, -2, 1],
          [1, 0, -4],
          [1, -10**6, 1];

for @sets -> @coefficients {

   say "Roots for @coefficients.join(', ').fmt("%-16s")",
       "=> (&quadroots( @coefficients ).join(', '))";

}

multi sub quadroots ($a, $b, $c) {

   my $root = (my $t = $b ** 2 - 4 * $a * $c ) < 0
       ?? $t.Complex.sqrt
       !! $t.sqrt;
   return ( -$b + $root ) / (2 * $a),
          ( -$b - $root ) / (2 * $a);

}

multi sub quadroots (@a) {

   @a == 3 or die "Expected three elements, got {+@a}";
   quadroots |@a;

}</lang> Output:

Roots for 1, 2, 1         => (-1, -1)
Roots for 1, 2, 3         => (-1 + 1.4142135623731i, -1 + -1.4142135623731i)
Roots for 1, -2, 1        => (1, 1)
Roots for 1, 0, -4        => (2, -2)
Roots for 1, -1000000, 1  => (999999.999999, 1.00000761449337e-06)

PL/I

<lang PL/I>

  declare (c1, c2) float complex,
          (a, b, c, x1, x2) float;

  get list (a, b, c);
  if b**2 < 4*a*c then
     do;
        c1 = (-b + sqrt(b**2 - 4+0i*a*c)) / (2*a);
        c2 = (-b - sqrt(b**2 - 4+0i*a*c)) / (2*a);
        put data (c1, c2);
     end;
  else
     do;
        x1 = (-b + sqrt(b**2 - 4*a*c)) / (2*a);
        x2 = (-b - sqrt(b**2 - 4*a*c)) / (2*a);
        put data (x1, x2);
     end;

</lang>

PicoLisp

<lang PicoLisp>(scl 40)

(de solveQuad (A B C)

  (let SD (sqrt (- (* B B) (* 4 A C)))
     (if (lt0 B)
        (list
           (*/ (- SD B) A 2.0)
           (*/ C 2.0 (*/ A A (- SD B) `(* 1.0 1.0))) )
        (list
           (*/ C 2.0 (*/ A A (- 0 B SD) `(* 1.0 1.0)))
           (*/ (- 0 B SD) A 2.0) ) ) ) )

(mapcar round

  (solveQuad 1.0 -1000000.0 1.0)
  (6 .) )</lang>

Output:

-> ("999,999.999999" "0.000001")

Python

<lang python>>>> def quad_discriminating_roots(a,b,c, entier = 1e-5): discriminant = b*b - 4*a*c a,b,c,d =complex(a), complex(b), complex(c), complex(discriminant) root1 = (-b + d**0.5)/2./a root2 = (-b - d**0.5)/2./a if abs(discriminant) < entier: return "real and equal", abs(root1), abs(root1) if discriminant > 0: return "real", root1.real, root2.real return "complex", root1, root2

>>> for coeffs in ((3, 4, 4/3.), (3, 2, -1), (3, 2, 1), (1.0, -10.0E5, 1.0)): print "Roots of: %gX^2 %+gX %+g are" % coeffs print " %s: %s, %s" % quad_discriminating_roots(*coeffs)

Roots of: 3X^2 +4X +1.33333 are

 real and equal: 0.666666666667, 0.666666666667

Roots of: 3X^2 +2X -1 are

 real: 0.333333333333, -1.0

Roots of: 3X^2 +2X +1 are

 complex: (-0.333333333333+0.471404520791j), (-0.333333333333-0.471404520791j)

Roots of: 1X^2 -1e+06X +1 are

 real: 999999.999999, 1.00000761449e-06

>>></lang>

R

Translation of: Python

<lang R>quaddiscrroots <- function(a,b,c, tol=1e-5) {

 d <- b*b - 4*a*c + 0i
 root1 <- (-b + sqrt(d))/(2*a)
 root2 <- (-b - sqrt(d))/(2*a)
 if ( abs(Re(d)) < tol ) {
   list("real and equal", abs(root1), abs(root1))
 } else if ( Re(d) > 0 ) {
   list("real", Re(root1), Re(root2))
 } else {
   list("complex", root1, root2)
 }

}

for(coeffs in list(c(3,4,4/3), c(3,2,-1), c(3,2,1), c(1, -1e6, 1)) ) {

 cat(sprintf("roots of %gx^2 %+gx^1 %+g are\n", coeffs[1], coeffs[2], coeffs[3]))
 r <- quaddiscrroots(coeffs[1], coeffs[2], coeffs[3])
 cat(sprintf("  %s: %s, %s\n", r1, r2, r3))

}</lang>

REXX

The REXX language has no SQRT function nor does it have complex numbers.
The SQRT function is included below.
Since unlimited decimal precision is part of the language, the NUMERIC DIGITS
was increased (from a default of 9) to 120 to accomidate roots closer to zero than the other. <lang rexx> /*REXX program finds the roots (may be complex) of a quadratic function.*/ numeric digits 120 /*use enough digits for extremes.*/ parse arg a b c . /*get specified arguments: A B C*/ a=a/1; b=b/1; c=c/1 /*normalize the three numbers. */ say 'a=' a /*show value of A. */ say 'b=' b /* " " " B. */ say 'c=' c /* " " " C. */ call quadratic a b c /*solve the quadratic function. */ numeric digits sqrt(digits())%1 /*reduce digits for human beans. */ r1=r1/1 /*normalize to the new digits. */ r2=r2/1 /* " " " " " */ if r1j\=0 then r1=r1||left('+',r1j>0)(r1j/1)"i" /*handle complex num.*/ if r2j\=0 then r2=r2||left('+',r2j>0)(r2j/1)"i" /* " " " */ say say 'root1=' r1 /*show 1st root (may be complex).*/ say 'root2=' r2 /* " 2nd " " " " */ exit

/*─────────────────────────────────────QUADRATIC subroutine─────────────*/ quadratic: parse arg aa bb cc . ?=sqrt(bb**2-4*aa*cc) /*compute sqrt (might be complex)*/ aa2=1/(aa+aa) /*compute reciprocal of 2*aa */ if right(?,1)=='i' then do /*are the roots complex? */

                       ?i=left(?,length(?)-1)
                       r1=-bb*aa2
                       r2=r1
                       r1j= ?i*aa2
                       r2j=-?i*aa2
                       end
                  else do
                       r1=(-bb+?)*aa2
                       r2=(-bb-?)*aa2
                       r1j=0
                       r2j=0
                       end

return

/*─────────────────────────────────────SQRT subroutine──────────────────*/ sqrt: procedure; parse arg x,f; if x=0 then return 0; d=digits(); numeric digits 11; g=sqrtguess();

 do j=0 while p>9; m.j=p; p=p%2+1; end;
 do k=j+5 to 0 by -1; if m.k>11 then numeric digits m.k; g=.5*(g+x/g); end;
 numeric digits d;return (g/1)i

sqrtguess: i=left('i',x<0); numeric form scientific; m.=11; p=d+d%4+2;

          parse value format(x,2,1,,0) 'E0' with g 'E' _ .; return g*.5'E'_%2

</lang> Output when using the input of:

1 -10e5 1

a= 1
b= -1000000
c= 1

root1= 1000000
root2= 0.000001

Output when using the input of:

1 -10e9 1

a= 1
b= -10000000000
c= 1

root1= 1.000000000E+10
root2= 1E-10

Output when using the input of:

3 2 1

a= 3
b= 2
c= 1

root1= -0.3333333333-0.5333870616i
root2= -0.3333333333+0.5333870616i

Ruby

With the

Library: complex.rb

package, the Math#sqrt method will return a Complex instance if necessary.

<lang ruby>require 'complex'

def quadratic(a, b, c)

 sqrt_discriminant = Math.sqrt(b**2 - 4*a*c)
 [(-b + sqrt_discriminant) / (2.0*a), (-b - sqrt_discriminant) / (2.0*a)]

end

p quadratic(3, 4, 4/3.0) # {-2/3} p quadratic(3, 2, -1) # {1/3, -1} p quadratic(3, 2, 1) # {(-1/3 + sqrt(2/9)i), (-1/3 - sqrt(2/9)i)} p quadratic(1, 0, 1) # {(0+i), (0-i)} p quadratic(1, -1e6, 1) # {1e6, 1e-6} p quadratic(-2, 7, 15) # {5, -3/2} p quadratic(1, -2, 1) # {1} p quadratic(1, 3, 3) # {(-3 - sqrt(3)i)/2), (-3 + sqrt(3)i)/2)}</lang>

[-0.666666666666667, -0.666666666666667]
[0.333333333333333, -1.0]
[Complex(-0.333333333333333, 0.471404520791032), Complex(-0.333333333333333, -0.471404520791032)]
[Complex(0.0, 1.0), Complex(0.0, -1.0)]
[999999.999999, 1.00000761449337e-06]
[-1.5, 5.0]
[1.0, 1.0]
[Complex(-1.5, 0.866025403784439), Complex(-1.5, -0.866025403784439)]

Scheme

<lang scheme>(define (quadratic a b c) (if (= a 0) (if (= b 0) 'fail (- (/ c b))) (let ((delta (- (* b b) (* 4 a c)))) (if (and (real? delta) (> delta 0)) (let ((u (+ b (* (if (>= b 0) 1 -1) (sqrt delta))))) (list (/ u -2 a) (/ (* -2 c) u))) (list (/ (- (sqrt delta) b) 2 a) (/ (+ (sqrt delta) b) -2 a))))))

examples

(quadratic 1 -1 -1)

(1.618033988749895 -0.6180339887498948)

(quadratic 1 0 -2)

(-1.4142135623730951 1.414213562373095)

(quadratic 1 0 2)

(0+1.4142135623730951i 0-1.4142135623730951i)

(quadratic 1+1i 2 5)

(-1.0922677260818898-1.1884256155834088i 0.09226772608188982+2.1884256155834088i)

(quadratic 0 4 3)

-3/4

(quadratic 0 0 1)

fail

(quadratic 1 2 0)

(-2 0)

(quadratic 1 2 1)

(-1 -1)

(quadratic 1 -1e5 1)

(99999.99999 1.0000000001000001e-05)</lang>

Tcl

Library: Tcllib (Package: math::complexnumbers)

<lang tcl>package require math::complexnumbers namespace import math::complexnumbers::complex math::complexnumbers::tostring

proc quadratic {a b c} {

   set discrim [expr {$b**2 - 4*$a*$c}]
   set roots [list]
   if {$discrim < 0} {
       set term1 [expr {(-1.0*$b)/(2*$a)}]
       set term2 [expr {sqrt(abs($discrim))/(2*$a)}]
       lappend roots [tostring [complex $term1 $term2]] \
               [tostring [complex $term1 [expr {-1 * $term2}]]]
   } elseif {$discrim == 0} {
       lappend roots [expr {-1.0*$b / (2*$a)}]
   } else {
       lappend roots [expr {(-1*$b + sqrt($discrim)) / (2 * $a)}] \
               [expr {(-1*$b - sqrt($discrim)) / (2 * $a)}]
   }
   return $roots

}

proc report_quad {a b c} {

   puts [format "%sx**2 + %sx + %s = 0" $a $b $c]
   foreach root [quadratic $a $b $c] {
       puts "    x = $root"
   }

}

1. examples on this page

report_quad 3 4 [expr {4/3.0}] ;# {-2/3} report_quad 3 2 -1  ;# {1/3, -1} report_quad 3 2 1  ;# {(-1/3 + sqrt(2/9)i), (-1/3 - sqrt(2/9)i)} report_quad 1 0 1  ;# {(0+i), (0-i)} report_quad 1 -1e6 1  ;# {1e6, 1e-6}

1. examples from http://en.wikipedia.org/wiki/Quadratic_equation

report_quad -2 7 15  ;# {5, -3/2} report_quad 1 -2 1  ;# {1} report_quad 1 3 3  ;# {(-3 - sqrt(3)i)/2), (-3 + sqrt(3)i)/2)}</lang> outputs

3x**2 + 4x + 1.3333333333333333 = 0
    x = -0.6666666666666666
3x**2 + 2x + -1 = 0
    x = 0.3333333333333333
    x = -1.0
3x**2 + 2x + 1 = 0
    x = -0.3333333333333333+0.47140452079103173i
    x = -0.3333333333333333-0.47140452079103173i
1x**2 + 0x + 1 = 0
    x = i
    x = -i
1x**2 + -1e6x + 1 = 0
    x = 999999.999999
    x = 1.00000761449337e-6
-2x**2 + 7x + 15 = 0
    x = -1.5
    x = 5.0
1x**2 + -2x + 1 = 0
    x = 1.0
1x**2 + 3x + 3 = 0
    x = -1.5+0.8660254037844386i
    x = -1.5-0.8660254037844386i

TI-89 BASIC

TI-89 BASIC has built-in numeric and algebraic solvers. solve(x^2-1E9x+1.0) returns x=1.E-9 or x=1.E9.