# 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 ${\displaystyle ax^{2}+bx+c=0}$. Your program must correctly handle non-real roots, but it need not check that ${\displaystyle a\neq 0}$.

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 ${\displaystyle a=1}$, ${\displaystyle b=-10^{5}}$, and ${\displaystyle c=1}$. (For double-precision floats, set ${\displaystyle b=-10^{9}}$.) Consider the following implementation in 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;
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 ${\displaystyle 10^{-6}}$. The naive method is numerically unstable.

Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters ${\displaystyle q={\sqrt {ac}}/b}$ and ${\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}$

and the two roots of the quardratic are: ${\displaystyle {\frac {-b}{a}}f}$ and ${\displaystyle {\frac {-c}{bf}}}$

Task: do it better. This means that given ${\displaystyle a=1}$, ${\displaystyle b=-10^{9}}$, and ${\displaystyle c=1}$, both of the roots your program returns should be greater than ${\displaystyle 10^{-11}}$. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle ${\displaystyle b=-10^{6}}$. 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.

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;

Here precision loss is prevented by checking signs of operands. On errors, Constraint_Error is propagated on numeric errors and when roots are complex.

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


## ALGOL 68

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
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); # 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  ODEND # quadratic_equation #
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

MsgBox % quadratic(u,v, 1,-3,2) ", " u ", " vMsgBox % quadratic(u,v, 1,3,2) ", " u ", " vMsgBox % quadratic(u,v, -2,4,-2) ", " u ", " vMsgBox % quadratic(u,v, 1,0,1) ", " u ", " vSetFormat FloatFast, 0.15eMsgBox % 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}

      FOR test% = 1 TO 7        READ a$, b$, c$PRINT "For a = " ; a$ ", b = " ; b$", c = " ; c$ TAB(32) ;        PROCsolvequadratic(EVAL(a$), EVAL(b$), EVAL(c$)) NEXT END DATA 1, -1E9, 1 DATA 1, 0, 1 DATA 2, -1, -6 DATA 1, 2, -2 DATA 0.5, SQR(2), 1 DATA 1, 3, 2 DATA 3, 4, 5 DEF PROCsolvequadratic(a, b, c) LOCAL d, f d = b^2 - 4*a*c CASE SGN(d) OF WHEN 0: PRINT "the single root is " ; -b/2/a WHEN +1: f = (1 + SQR(1-4*a*c/b^2))/2 PRINT "the real roots are " ; -f*b/a " and " ; -c/b/f WHEN -1: PRINT "the complex roots are " ; -b/2/a " +/- " ; SQR(-d)/2/a "*i" ENDCASE ENDPROC Output: For a = 1, b = -1E9, c = 1 the real roots are 1E9 and 1E-9 For a = 1, b = 0, c = 1 the complex roots are 0 +/- 1*i For a = 2, b = -1, c = -6 the real roots are 2 and -1.5 For a = 1, b = 2, c = -2 the real roots are -2.73205081 and 0.732050808 For a = 0.5, b = SQR(2), c = 1 the single root is -1.41421356 For a = 1, b = 3, c = 2 the real roots are -2 and -1 For a = 3, b = 4, c = 5 the complex roots are -0.666666667 +/- 1.1055416*i ## C Code that tries to avoid floating point overflow and other unfortunate loss of precissions: (compiled with gcc -std=c99 for complex, though easily adapted to just real numbers) #include <stdio.h>#include <stdlib.h>#include <complex.h>#include <math.h> typedef double complex cplx; void quad_root(double a, double b, double c, cplx * ra, cplx *rb){ double d, e; if (!a) { *ra = b ? -c / b : 0; *rb = 0; return; } if (!c) { *ra = 0; *rb = -b / a; return; } b /= 2; if (fabs(b) > fabs(c)) { e = 1 - (a / b) * (c / b); d = sqrt(fabs(e)) * fabs(b); } else { e = (c > 0) ? a : -a; e = b * (b / fabs(c)) - e; d = sqrt(fabs(e)) * sqrt(fabs(c)); } if (e < 0) { e = fabs(d / a); d = -b / a; *ra = d + I * e; *rb = d - I * e; return; } d = (b >= 0) ? d : -d; e = (d - b) / a; d = e ? (c / e) / a : 0; *ra = d; *rb = e; return;} int main(){ cplx ra, rb; quad_root(1, 1e12 + 1, 1e12, &ra, &rb); printf("(%g + %g i), (%g + %g i)\n", creal(ra), cimag(ra), creal(rb), cimag(rb)); quad_root(1e300, -1e307 + 1, 1e300, &ra, &rb); printf("(%g + %g i), (%g + %g i)\n", creal(ra), cimag(ra), creal(rb), cimag(rb)); return 0;} Output: (-1e+12 + 0 i), (-1 + 0 i) (1.00208e+07 + 0 i), (9.9792e-08 + 0 i) #include <stdio.h>#include <math.h>#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);} Translation of: 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; }} 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;} 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# 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)); }} Output: ((1E+20, 0), (1E-20, 0)) ## C++ #include <iostream>#include <utility>#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;} Output: (1e+20,0), (1e-20,0)  ## Clojure (defn quadratic "Compute the roots of a quadratic in the form ax^2 + bx + c = 1. Returns any of nil, a float, or a vector." [a b c] (let [sq-d (Math/sqrt (- (* b b) (* 4 a c))) f #(/ (% b sq-d) (* 2 a))] (cond (neg? sq-d) nil (zero? sq-d) (f +) (pos? sq-d) [(f +) (f -)] :else nil))) ; maybe our number ended up as NaN Output: user=> (quadratic 1.0 1.0 1.0)niluser=> (quadratic 1.0 2.0 1.0)1.0user=> (quadratic 1.0 3.0 1.0)[2.618033988749895 0.3819660112501051]  ## Common Lisp (defun quadratic (a b c) (list (/ (+ (- b) (sqrt (- (expt b 2) (* 4 a c)))) (* 2 a)) (/ (- (- b) (sqrt (- (expt b 2) (* 4 a c)))) (* 2 a)))) ## D import std.math, std.traits; CommonType!(T1, T2, T3)[] naiveQR(T1, T2, T3) (in T1 a, in T2 b, in T3 c)pure nothrow if (isFloatingPoint!T1) { alias ReturnT = typeof(typeof(return).init[0]); if (a == 0) return [ReturnT(c / b)]; // It's a linear function. immutable ReturnT det = b ^^ 2 - 4 * a * c; if (det < 0) return []; // No real number root. immutable SD = sqrt(det); return [(-b + SD) / 2 * a, (-b - SD) / 2 * a];} CommonType!(T1, T2, T3)[] cautiQR(T1, T2, T3) (in T1 a, in T2 b, in T3 c)pure nothrow if (isFloatingPoint!T1) { alias ReturnT = typeof(typeof(return).init[0]); if (a == 0) return [ReturnT(c / b)]; // It's a linear function. immutable ReturnT det = b ^^ 2 - 4 * a * c; if (det < 0) return []; // No real number root. immutable SD = sqrt(det); if (b * a < 0) { immutable x = (-b + SD) / 2 * a; return [x, c / (a * x)]; } else { immutable x = (-b - SD) / 2 * a; return [c / (a * x), x]; }} void main() { import std.stdio; writeln("With 32 bit float type:"); writefln(" Naive: [%(%g, %)]", naiveQR(1.0f, -10e5f, 1.0f)); writefln("Cautious: [%(%g, %)]", cautiQR(1.0f, -10e5f, 1.0f)); writeln("\nWith 64 bit double type:"); writefln(" Naive: [%(%g, %)]", naiveQR(1.0, -10e5, 1.0)); writefln("Cautious: [%(%g, %)]", cautiQR(1.0, -10e5, 1.0)); writeln("\nWith real type:"); writefln(" Naive: [%(%g, %)]", naiveQR(1.0L, -10e5L, 1.0L)); writefln("Cautious: [%(%g, %)]", cautiQR(1.0L, -10e5L, 1.0L));} Output: With 32 bit float type: Naive: [1e+06, 0] Cautious: [1e+06, 1e-06] With 64 bit double type: Naive: [1e+06, 1.00001e-06] Cautious: [1e+06, 1e-06] With real type: Naive: [1e+06, 1e-06] Cautious: [1e+06, 1e-06] ## Elixir defmodule Quadratic do def roots(a, b, c) do IO.puts "Roots of a quadratic function (#{a}, #{b}, #{c})" d = b * b - 4 * a * c a2 = a * 2 cond do d > 0 -> sd = :math.sqrt(d) IO.puts " the real roots are #{(- b + sd) / a2} and #{(- b - sd) / a2}" d == 0 -> IO.puts " the single root is #{- b / a2}" true -> sd = :math.sqrt(-d) IO.puts " the complex roots are #{- b / a2} +/- #{sd / a2}*i" end endend Quadratic.roots(1, -2, 1)Quadratic.roots(1, -3, 2)Quadratic.roots(1, 0, 1)Quadratic.roots(1, -1.0e10, 1)Quadratic.roots(1, 2, 3)Quadratic.roots(2, -1, -6) Output: Roots of a quadratic function (1, -2, 1) the single root is 1.0 Roots of a quadratic function (1, -3, 2) the real roots are 2.0 and 1.0 Roots of a quadratic function (1, 0, 1) the complex roots are 0.0 +/- 1.0*i Roots of a quadratic function (1, -1.0e10, 1) the real roots are 1.0e10 and 0.0 Roots of a quadratic function (1, 2, 3) the complex roots are -1.0 +/- 1.4142135623730951*i Roots of a quadratic function (2, -1, -6) the real roots are 2.0 and -1.5  ## ERRE PROGRAM QUADRATIC PROCEDURE SOLVE_QUADRATIC D=B*B-4*A*C IF ABS(D)<1D-6 THEN D=0 END IF CASE SGN(D) OF 0-> PRINT("the single root is ";-B/2/A) END -> 1-> F=(1+SQR(1-4*A*C/(B*B)))/2 PRINT("the real roots are ";-F*B/A;"and ";-C/B/F) END -> -1-> PRINT("the complex roots are ";-B/2/A;"+/-";SQR(-D)/2/A;"*i") END -> END CASEEND PROCEDURE BEGIN PRINT(CHR$(12);) ! CLS  FOR TEST%=1 TO 7 DO     READ(A,B,C)     PRINT("For a=";A;",b=";B;",c=";C;TAB(32);)     SOLVE_QUADRATIC  END FOR  DATA(1,-1E9,1)  DATA(1,0,1)  DATA(2,-1,-6)  DATA(1,2,-2)  DATA(0.5,1.4142135,1)  DATA(1,3,2)  DATA(3,4,5)END PROGRAM
Output:
For a= 1 ,b=-1E+09 ,c= 1       the real roots are  1E+09 and  1E-09
For a= 1 ,b= 0 ,c= 1           the complex roots are  0 +/- 1 *i
For a= 2 ,b=-1 ,c=-6           the real roots are  2 and -1.5
For a= 1 ,b= 2 ,c=-2           the real roots are -2.732051 and  .7320508
For a= .5 ,b= 1.414214 ,c= 1   the single root is -1.414214
For a= 1 ,b= 3 ,c= 2           the real roots are -2 and -1
For a= 3 ,b= 4 ,c= 5           the complex roots are -.6666667 +/- 1.105542 *i


## 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 * / ;
( scratchpad ) 1 -1.e20 1 quadratic-equation--- Data stack:1.0e+209.999999999999999e-21

Middlebrook method

:: quadratic-equation2 ( a b c -- x1 x2 ) a c * sqrt b / :> q  1 4 q sq * - sqrt 0.5 * 0.5 + :> f b neg a / f * c neg b / f / ;

( scratchpad ) 1 -1.e20 1 quadratic-equation--- Data stack:1.0e+201.0e-20

## Forth

Without locals:

: 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/ ;

With locals:

: 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/ ; \ test1 set-precision1e -1e6 1e quadratic fs. fs.     \ 1e-6 1e6

## Fortran

### Fortran 90

Works with: Fortran version 90 and later
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    rroot1 = -(b + SIGN(SQRT(discriminant), b)) / (2.0_dp * a)    rroot2 = c / (a * rroot1)    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
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


### Fortran I

Source code written in FORTRAN I (october 1956) for the IBM 704.

 COMPUTE ROOTS OF A QUADRATIC FUNCTION - 1956      READ 100,A,B,C 100  FORMAT(3F8.3)      PRINT 100,A,B,C      DISC=B**2-4.*A*C      IF(DISC),1,2,3 1    XR=-B/(2.*A)      XI=SQRT(-DISC)/(2.*A)      XJ=-XI      PRINT 311      PRINT 312,XR,XI,XR,XJ 311  FORMAT(13HCOMPLEX ROOTS) 312  FORMAT(4HX1=(,2E12.4,6H),X2=(,2E12.4,1H))      GO TO 999 2    X1=-B/(2.*A)      X2=X1      PRINT 321      PRINT 332,X1,X2 321  FORMAT(16HEQUAL REAL ROOTS)      GO TO 999 3    X1= (-B+SQRT(DISC)) / (2.*A)      X2= (-B-SQRT(DISC)) / (2.*A)      PRINT 331      PRINT 332,X1,X2 331  FORMAT(10HREAL ROOTS) 332  FORMAT(3HX1=,E12.5,4H,X2=,E12.5) 999  STOP

## 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; # Hint : E(12) is a 12th primitive root of 1QuadraticRoots(2, 2, -1);# [ 1/2*E(12)^4-1/2*E(12)^7+1/2*E(12)^8+1/2*E(12)^11,#   1/2*E(12)^4+1/2*E(12)^7+1/2*E(12)^8-1/2*E(12)^11 ] # This works also with floating-point numbersQuadraticRoots(2.0, 2.0, -1.0);# [ 0.366025, -1.36603 ]

## 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() {    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])    }}
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


## J

Solution use J's built-in polynomial solver:

   p.


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

   coeff =. _3 |.\ 3 4 4r3   3 2 _1   3 2 1   1 _1e6 1   1 _1e9 1   > {:"1 p. coeff         _0.666667           _0.666667                _1            0.333333_0.333333j0.471405 _0.333333j_0.471405               1e6                1e_6               1e9                1e_9

Of course p. generalizes to polynomials of arbitrary order (which isn't as great as that might sound, because of practical limitations). 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 ${\displaystyle 0+16x-12x^{2}+2x^{3}}$:

   p. 0 16 _12 2   NB. return multiplier ; roots+-+-----+|2|4 2 0|+-+-----+   p. 2 ; 4 2 0    NB. return coefficients0 16 _12 2

Exploring the limits of precision:

   1{::p. 1 _1e5 1                   NB. display roots100000 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 roots1 9.99999e_11   1e5 1e_5 - 1{::p. 1 _1e5 1        NB. find difference1e_5 _1e_15   1 _1e5 1 p. 1e5 1e_5-1e_5 _1e_15  NB. test displayed roots with adjustment_3.38436e_7 0

When these "roots" are plugged back into the original polynomial, the results are nowhere near zero. However, double precision floating point does not have enough bits to represent the (extremely close) answers that would give a zero result.

Middlebrook formula implemented in J

q_r=: verb define  'a b c' =. y  q=. b %~ %: a * c  f=. 0.5 + 0.5 * %:(1-4*q*q)  (-b*f%a),(-c%b*f))    q_r 1 _1e6 11e6 1e_6

## Java

public class QuadraticRoots {    private 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;  // discriminant        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 if (b < 0.0) {            // Avoid calculating -b - Math.sqrt(d), to avoid any            // subtractive cancellation when it is near zero.            double re = (-b + Math.sqrt(d)) / aa;            roots[0] = new Complex(re, 0.0);            roots[1] = new Complex(c / (a * re), 0.0);        } else {            // Avoid calculating -b + Math.sqrt(d).            double re = (-b - Math.sqrt(d)) / aa;            roots[1] = new Complex(re, 0.0);            roots[0] = new Complex(c / (a * re), 0.0);        }        return roots;    }     public static void main(String[] args) {        double[][] equations = {            {1.0, 22.0, -1323.0},   // two distinct real roots            {6.0, -23.0, 20.0},     //   with a != 1.0            {1.0, -1.0e9, 1.0},     //   with one root near zero            {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]);            }        }    }}
Output:
a = 1.00000   b = 22.0000   c = -1323.00
X1 = 27.0000
X2 = -49.0000

a = 6.00000   b = -23.0000   c = 20.0000
X1 = 2.50000
X2 = 1.33333

a = 1.00000   b = -1.00000e+09   c = 1.00000
X1 = 1.00000e+09
X2 = 1.00000e-09

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

## jq

Works with: jq version 1.4

Currently jq does not include support for complex number operations, so a small library is included in the first section.

The second section defines quadratic_roots(a;b;c), which emits a stream of 0 or two solutions, or the value true if a==b==c==0.

The third section defines a function for producing a table showing (i, error, solution) for solutions to x^2 - 10^i + 1 = 0 for various values of i.

Section 1: Complex numbers (scrolling window)

# Complex numbers as points [x,y] in the Cartesian planedef real(z): if (z|type) == "number" then z else z[0] end; def imag(z): if (z|type) == "number" then 0 else z[1] end; def plus(x; y):    if (x|type) == "number" then       if  (y|type) == "number" then [ x+y, 0 ]       else [ x + y[0], y[1]]       end    elif (y|type) == "number" then plus(y;x)    else [ x[0] + y[0], x[1] + y[1] ]    end; def multiply(x; y):    if (x|type) == "number" then       if  (y|type) == "number" then [ x*y, 0 ]       else [x * y[0], x * y[1]]       end    elif (y|type) == "number" then multiply(y;x)    else [ x[0] * y[0] - x[1] * y[1],  x[0] * y[1] + x[1] * y[0]]    end; def negate(x): multiply(-1; x); def minus(x; y): plus(x; multiply(-1; y)); def conjugate(z):  if (z|type) == "number" then [z, 0]  else  [z[0], -(z[1]) ]  end; def invert(z):  if (z|type) == "number" then [1/z, 0]  else    ( (z[0] * z[0]) + (z[1] * z[1]) ) as $d # use "0 + ." to convert -0 back to 0 | [ z[0]/$d, (0 + -(z[1]) / $d)] end; def divide(x;y): multiply(x; invert(y)); def magnitude(z): real( multiply(z; conjugate(z))) | sqrt; # exp^zdef complex_exp(z): def expi(x): [ (x|cos), (x|sin) ]; if (z|type) == "number" then z|exp elif z[0] == 0 then expi(z[1]) # for efficiency else multiply( (z[0]|exp); expi(z[1]) ) end ; def complex_sqrt(z): if imag(z) == 0 and real(z) >= 0 then [(real(z)|sqrt), 0] else magnitude(z) as$r    | if $r == 0 then [0,0] else (real(z)/$r) as $a | (imag(z)/$r) as $b |$r | sqrt as $r | (($a | acos) / 2)       | [ ($r * cos), ($r * sin)]      end  end ;

# If there are infinitely many solutions, emit true;# if none, emit empty; # otherwise always emit two.# For numerical accuracy, Middlebrook's approach is adopted:def quadratic_roots(a; b; c):  if a == 0 and b == 0 then     if c == 0 then true # infinitely many     else empty          # none     end  elif a == 0 then [-c/b, 0]  elif b == 0 then (complex_sqrt(1/a) | (., negate(.)))  else    divide( plus(1.0; complex_sqrt( minus(1.0; (4 * a * c / (b*b))))); 2) as $f | negate(divide(multiply(b;$f); a)),      negate(divide(c; multiply(b; $f))) end; Section 3: Produce a table showing [i, error, solution] for solutions to x^2 - 10^i + 1 = 0 def example: def pow(i): . as$in | reduce range(0;i) as $i (1; . *$in);  def poly(a;b;c): plus( plus( multiply(a; multiply(.;.)); multiply(b;.)); c);  def abs: if . < 0 then -. else . end;  def zero(z):    if z == 0 then 0    else (magnitude(z)|abs) as $zero | if$zero < 1e-10 then "+0" else $zero end end; def lpad(n): tostring | (n - length) * " " + .; range(0; 13) as$i  | (- (10|pow($i))) as$b  | quadratic_roots(1; $b; 1) as$x  | $x | poly(1;$b; 1) as $zero | "\($i|lpad(4)): error = \(zero($zero)|lpad(18)) x=\($x)" ; example
Output:
(scrolling window)
 $jq -M -r -n -f Roots_of_a_quadratic_function.jq 0: error = +0 x=[0.5,0.8660254037844386] 0: error = +0 x=[0.5000000000000001,-0.8660254037844387] 1: error = +0 x=[9.898979485566356,0] 1: error = +0 x=[0.10102051443364382,-0] 2: error = +0 x=[99.98999899979995,0] 2: error = +0 x=[0.010001000200050014,-0] 3: error = 1.1641532182693481e-10 x=[999.998999999,0] 3: error = +0 x=[0.0010000010000019998,-0] 4: error = +0 x=[9999.999899999999,0] 4: error = +0 x=[0.00010000000100000003,-0] 5: error = +0 x=[99999.99999,0] 5: error = +0 x=[1.0000000001e-05,-0] 6: error = 0.0001220703125 x=[999999.9999989999,0] 6: error = +0 x=[1.000000000001e-06,-0] 7: error = 0.015625 x=[9999999.9999999,0] 7: error = +0 x=[1.0000000000000101e-07,-0] 8: error = 1 x=[99999999.99999999,0] 8: error = +0 x=[1e-08,-0] 9: error = 1 x=[1000000000,0] 9: error = +0 x=[1e-09,-0] 10: error = 1 x=[10000000000,0] 10: error = +0 x=[1e-10,-0] 11: error = 1 x=[100000000000,0] 11: error = +0 x=[1e-11,-0] 12: error = 1 x=[1000000000000,0] 12: error = +0 x=[1e-12,-0] ## Julia This solution is an implementation of algorithm from the Goldberg paper cited in the task description. It does check for a=0 and returns the linear solution in that case. Julia's sqrt throws a domain error for negative real inputs, so negative discriminants are converted to complex by adding 0im prior to taking the square root. Alternative solutions might make use of Julia's Polynomials or Roots packages. function quadroots(x::Real, y::Real, z::Real) a, b, c = promote(float(x), y, z) if a ≈ 0.0 return [-c / b] end Δ = b ^ 2 - 4a * c if Δ ≈ 0.0 return [-sqrt(c / a)] end if Δ < 0.0 Δ = complex(Δ) end d = sqrt(Δ) if b < 0.0 d -= b return [d / 2a, 2c / d] else d = -d - b return [2c / d, d / 2a] endend a = [1, 1, 1.0, 10]b = [10, 2, -10.0 ^ 9, 1]c = [1, 1, 1, 1] for (x, y, z) in zip(a, b, c) @printf "The roots of %.2fx² + %.2fx + %.2f\n\tx₀ = (%s)\n" x y z join(round.(quadroots(x, y, z), 2), ", ")end Output: The roots of 1.00x² + 10.00x + 1.00 x₀ = (-0.1, -9.9) The roots of 1.00x² + 2.00x + 1.00 x₀ = (-1.0) The roots of 1.00x² + -1000000000.00x + 1.00 x₀ = (1.0e9, 0.0) The roots of 10.00x² + 1.00x + 1.00 x₀ = (-0.05 + 0.31im, -0.05 - 0.31im) ## Kotlin Translation of: Java import java.lang.Math.* data class Equation(val a: Double, val b: Double, val c: Double) { data class Complex(val r: Double, val i: Double) { override fun toString() = when { i == 0.0 -> r.toString() r == 0.0 -> "${i}i"            else -> "$r +${i}i"        }    }     data class Solution(val x1: Any, val x2: Any) {        override fun toString() = when(x1) {            x2 -> "X1,2 = $x1" else -> "X1 =$x1, X2 = $x2" } } val quadraticRoots by lazy { val _2a = a + a val d = b * b - 4.0 * a * c // discriminant if (d < 0.0) { val r = -b / _2a val i = sqrt(-d) / _2a Solution(Complex(r, i), Complex(r, -i)) } else { // avoid calculating -b +/- sqrt(d), to avoid any // subtractive cancellation when it is near zero. val r = if (b < 0.0) (-b + sqrt(d)) / _2a else (-b - sqrt(d)) / _2a Solution(r, c / (a * r)) } }} fun main(args: Array<String>) { val equations = listOf(Equation(1.0, 22.0, -1323.0), // two distinct real roots Equation(6.0, -23.0, 20.0), // with a != 1.0 Equation(1.0, -1.0e9, 1.0), // with one root near zero Equation(1.0, 2.0, 1.0), // one real root (double root) Equation(1.0, 0.0, 1.0), // two imaginary roots Equation(1.0, 1.0, 1.0)) // two complex roots equations.forEach { println("$it\n" + it.quadraticRoots) }}
Output:
Equation(a=1.0, b=22.0, c=-1323.0)
X1 = -49.0, X2 = 27.0
Equation(a=6.0, b=-23.0, c=20.0)
X1 = 2.5, X2 = 1.3333333333333333
Equation(a=1.0, b=-1.0E9, c=1.0)
X1 = 1.0E9, X2 = 1.0E-9
Equation(a=1.0, b=2.0, c=1.0)
X1,2 = -1.0
Equation(a=1.0, b=0.0, c=1.0)
X1 = 1.0i, X2 = -1.0i
Equation(a=1.0, b=1.0, c=1.0)
X1 = -0.5 + 0.8660254037844386i, X2 = -0.5 + -0.8660254037844386i

## lambdatalk

 1) using lambdas: {def equation {lambda {:a :b :c}  {b equation :a*x{sup 2}+:b*x+:c=0}  {{lambda {:a' :b' :d}   {if {> :d 0}   then {{lambda {:b' :d'}                   {equation.disp {+ :b' :d'} {- :b' :d'} 2 real roots}        } :b' {/ {sqrt :d} :a'}}   else {if {< :d 0}   then {{lambda {:b' :d'}          {equation.disp [:b',:d'] [:b',-:d'] 2 complex roots}        } :b' {/ {sqrt {- :d}} :a'} }   else {equation.disp :b'  :b' one real double root}  }}  } {* 2 :a} {/ {- :b} {* 2 :a}} {- {* :b :b} {* 4 :a :c}} } }}  2) using let: {def equation {lambda {:a :b :c}  {b equation :a*x{sup 2}+:b*x+:c=0}  {let { {:a' {* 2 :a}}         {:b' {/ {- :b} {* 2 :a}}}         {:d  {- {* :b :b} {* 4 :a :c}}} }   {if {> :d 0}    then {let { {:b' :b'}                {:d' {/ {sqrt :d} :a'}} }          {equation.disp {+ :b' :d'} {- :b' :d'} 2 real roots} }    else {if {< :d 0}    then {let { {:b' :b'}                {:d' {/ {sqrt {- :d}} :a'}} }           {equation.disp [:b',:d'] [:b',-:d'] 2 complex roots} }     else  {equation.disp :b' :b' one real double root} }} }}} 3) a function to display results in an HTML table format {def equation.disp {lambda {:x1 :x2 :txt}  {table {@ style="background:#ffa"}    {tr {td :txt:    }}   {tr {td x1 = :x1 }}    {tr {td x2 = :x2 }} } }} 4) testing: equation 1*x2+1*x+-1=0 2 real roots: x1 = 0.6180339887498949 x2 = -1.618033988749895 equation 1*x2+1*x+1=0 2 complex roots: x1 = [-0.5,0.8660254037844386] x2 = [-0.5,-0.8660254037844386] equation 1*x2+-2*x+1=0 one real double root: x1 = 1 x2 = 1

## Liberty BASIC

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 ifend function

## 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

## 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:

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 rootend for i = 1, 12 do  print(qsolve(1, 0-10^i, 1))end

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.

## Maple

solve(a*x^2+b*x+c,x); solve(1.0*x^2-10.0^9*x+1.0,x,explicit,allsolutions); fsolve(x^2-10^9*x+1,x,complex);
Output:
                                (1/2)                     (1/2)
/          2\             /          2\
-b + \-4 a c + b /         b + \-4 a c + b /
-----------------------, - ----------------------
2 a                       2 a

9                -9
1.000000000 10 , 1.000000000 10

-9                9
1.000000000 10  , 1.000000000 10 

## Mathematica

Possible ways to do this are (symbolic and numeric examples):

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}]

gives back:

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

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

${\displaystyle 50000-{\sqrt {2499999999}}}$

${\displaystyle 50000+{\sqrt {2499999999}}}$

{\displaystyle {\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}}}

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

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

${\displaystyle \{x\to 0.00001\}}$

${\displaystyle \{x\to 100000.\}}$

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 / Octave

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

## Perl 6

Perl 6 has complex number handling built in.

for[1, 2, 1],[1, 2, 3],[1, -2, 1],[1, 0, -4],[1, -10**6, 1]-> @coefficients {    printf "Roots for %d, %d, %d\t=> (%s, %s)\n",    |@coefficients, |quadroots(@coefficients);} sub quadroots (*[$a,$b, $c]) { ( -$b + $_ ) / (2 *$a),    ( -$b -$_ ) / (2 * $a) given ($b ** 2 - 4 * $a *$c ).Complex.sqrt.narrow}
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)

## Phix

Translation of: ERRE
procedure solve_quadratic(sequence t3)atom {a,b,c} = t3atom d = b*b-4*a*c, fstring s = sprintf("for a=%g,b=%g,c=%g",t3), tsequence u    if abs(d)<1e-6 then d=0 end if    switch sign(d) do        case 0: t = "single root is %g"                u = {-b/2/a}        case 1: t = "real roots are %g and %g"                f = (1+sqrt(1-4*a*c/(b*b)))/2                u = {-f*b/a,-c/b/f}        case-1: t = "complex roots are %g +/- %g*i"                u = {-b/2/a,sqrt(-d)/2/a}    end switch    printf(1,"%-25s the %s\n",{s,sprintf(t,u)})end procedure constant tests = {{1,-1E9,1},                  {1,0,1},                  {2,-1,-6},                  {1,2,-2},                  {0.5,1.4142135,1},                  {1,3,2},                  {3,4,5}} for i=1 to length(tests) do    solve_quadratic(tests[i])end for
for a=1,b=-1e+9,c=1       the real roots are 1e+9 and 1e-9
for a=1,b=0,c=1           the complex roots are 0 +/- 1*i
for a=2,b=-1,c=-6         the real roots are 2 and -1.5
for a=1,b=2,c=-2          the real roots are -2.73205 and 0.732051
for a=0.5,b=1.41421,c=1   the single root is -1.41421
for a=1,b=3,c=2           the real roots are -2 and -1
for a=3,b=4,c=5           the complex roots are -0.666667 +/- 1.10554*i


## 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 .) )
Output:
-> ("999,999.999999" "0.000001")

## 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;

## Python

Library: NumPy

This solution compares the naïve method with three "better" methods.

#!/usr/bin/env python3 import mathimport cmathimport numpy def quad_discriminating_roots(a,b,c, entier = 1e-5):    """For reference, the naive algorithm which shows complete loss of    precision on the quadratic in question.  (This function also returns a    characterization of the roots.)"""    discriminant = b*b - 4*a*c    a,b,c,d =complex(a), complex(b), complex(c), complex(discriminant)    root1 = (-b + cmath.sqrt(d))/2./a    root2 = (-b - cmath.sqrt(d))/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 def middlebrook(a, b, c):    try:        q = math.sqrt(a*c)/b        f = .5+ math.sqrt(1-4*q*q)/2    except ValueError:        q = cmath.sqrt(a*c)/b        f = .5+ cmath.sqrt(1-4*q*q)/2    return (-b/a)*f, -c/(b*f) def whatevery(a, b, c):    try:        d = math.sqrt(b*b-4*a*c)    except ValueError:        d = cmath.sqrt(b*b-4*a*c)    if b > 0:        return div(2*c, (-b-d)), div((-b-d), 2*a)    else:        return div((-b+d), 2*a), div(2*c, (-b+d)) def div(n, d):    """Divide, with a useful interpretation of division by zero."""    try:        return n/d    except ZeroDivisionError:        if n:            return n*float('inf')        return float('nan') testcases = [    (3, 4, 4/3),    # real, equal    (3, 2, -1),     # real, unequal    (3, 2, 1),      # complex    (1, -1e9, 1),   # ill-conditioned "quadratic in question" required by task.    (1, -1e100, 1),    (1, -1e200, 1),    (1, -1e300, 1),] print('Naive:')for c in testcases:    print("{} {:.5} {:.5}".format(*quad_discriminating_roots(*c))) print('\nMiddlebrook:')for c in testcases:    print(("{:.5} "*2).format(*middlebrook(*c))) print('\nWhat Every...')for c in testcases:    print(("{:.5} "*2).format(*whatevery(*c))) print('\nNumpy:')for c in testcases:    print(("{:.5} "*2).format(*numpy.roots(c)))
Output:
Naive:
real and equal 0.66667 0.66667
real 0.33333 -1.0
complex (-0.33333+0.4714j) (-0.33333-0.4714j)
real 1e+09 0.0
real 1e+100 0.0
real nan nan
real nan nan

Middlebrook:
-0.66667 -0.66667
(-1+0j) (0.33333+0j)
(-0.33333-0.4714j) (-0.33333+0.4714j)
1e+09 1e-09
1e+100 1e-100
1e+200 1e-200
1e+300 1e-300

What Every...
-0.66667 -0.66667
0.33333 -1.0
(-0.33333+0.4714j) (-0.33333-0.4714j)
1e+09 1e-09
1e+100 1e-100
inf 0.0
inf 0.0

Numpy:
-0.66667 -0.66667
-1.0 0.33333
(-0.33333+0.4714j) (-0.33333-0.4714j)
1e+09 1e-09
1e+100 1e-100
1e+200 1e-200
1e+300 0.0


## R

Translation of: Python
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", r[[1]], r[[2]], r[[3]]))}

## Racket

#lang racket(define (quadratic a b c)  (let* ((-b (- b))         (delta (- (expt b 2) (* 4 a c)))         (denominator (* 2 a)))    (list     (/ (+ -b (sqrt delta)) denominator)     (/ (- -b (sqrt delta)) denominator)))) ;(quadratic 1 0.0000000000001 -1);'(0.99999999999995 -1.00000000000005);(quadratic 1 0.0000000000001 1);'(-5e-014+1.0i -5e-014-1.0i)

## REXX

### version 1

The REXX language doesn't have a   sqrt   function,   nor does it support complex numbers natively.

Since "unlimited" decimal precision is part of the REXX language, the   numeric digits   was increased
(from a default of   9)   to   200   to accommodate when a root is closer to zero than the other root.

Note that only ten decimal digits (precision) are shown in the   displaying   of the output.

This REXX version supports   complex numbers   for the result.

/*REXX program finds the roots (which may be complex) of a quadratic function.*/numeric digits 200                     /*use enough digits to handle extremes.*/parse arg a b c .                      /*obtain the specified arguments: A B C*/call quadratic  a,b,c                  /*solve quadratic function via the sub.*/numeric digits  10                     /*reduce (output) digits for human eyes*/r1=r1/1;  r2=r2/1; a=a/1; b=b/1; c=c/1 /*normalize numbers to the new digits. */if r1j\=0  then r1=r1 || left('+',r1j>0)(r1j/1)"i"    /*handle complex number.*/if r2j\=0  then r2=r2 || left('+',r2j>0)(r2j/1)"i"    /*   "      "       "   */              say '    a ='   a        /*display the normalized value of   A. */              say '    b ='   b        /*   "     "       "       "    "   B. */              say '    c ='   c        /*   "     "       "       "    "   C. */      say;    say 'root1 ='   r1       /*   "     "       "       "   1st root*/              say 'root2 ='   r2       /*   "     "       "       "   2nd root*/exit                                   /*stick a fork in it,  we're all done. *//*────────────────────────────────────────────────────────────────────────────*/quadratic:  parse arg aa,bb,cc         /*obtain the specified three arguments.*/   $=sqrt(bb**2-4*aa*cc); L=length($) /*compute  SQRT (which may be complex).*/   r=1/(aa+aa);     ?=right($,1)=='i' /*compute reciprocal of 2*aa; Complex?*/ if ? then do; r1= -bb *r; r2=r1; r1j=left($,L-1)*r; r2j=-r1j; end         else do;  r1=(-bb+$)*r; r2=(-bb-$)*r; r1j=0;             r2j= 0;   end   return/*────────────────────────────────────────────────────────────────────────────*/sqrt: procedure; parse arg x 1 ox;   if x=0  then return 0;   d=digits();   m.=9      numeric digits 9;    numeric form;    h=d+6;    x=abs(x)      parse value format(x,2,1,,0)  'E0'   with   g 'E' _ .;      g=g *.5'e'_ %2            do j=0  while h>9;      m.j=h;              h=h%2+1;       end /*j*/            do k=j+5  to 0  by -1;  numeric digits m.k; g=(g+x/g)*.5;  end /*k*/      numeric digits d;    return (g/1)left('i',ox<0)  /*make complex if OX<0.*/

output   when using the input of:   1   -10e5   1

    a = 1
b = -1000000
c = 1

root1 = 1000000
root2 = 0.000001


The following output is when Regina 3.9.1 REXX is used.

output   when using the input of:   1   -10e9   1

    a = 1
b = -10000000000
c = 1

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


The following output is when R4 REXX is used.

output   when using the input of:   1   -10e9   1

    a = 1
b = -1E+10
c = 1

root1 = 1E+10
root2 = 0.0000000001


output   when using the input of:   3   2   1

    a = 3
b = 2
c = 1

root1 = -0.3333333333+0.4714045208i
root2 = -0.3333333333-0.4714045208i


output   when using the input of:   1   0   1

    a = 1
b = 0
c = 1

root1 = 0+1i
root2 = 0-1i


### Version 2

/* REXX **************************************************************** 26.07.2913 Walter Pachl**********************************************************************/  Numeric Digits 30  Parse Arg a b c 1 alist  Select    When a='' | a='?' Then      Call exit 'rexx qgl a b c solves a*x**2+b*x+c'    When words(alist)<>3 Then      Call exit 'three numbers are required'    Otherwise      Nop    End  gl=a'*x**2'  Select    When b<0 Then gl=gl||b'*x'    When b>0 Then gl=gl||'+'||b'*x'    Otherwise Nop    End  Select    When c<0 Then gl=gl||c    When c>0 Then gl=gl||'+'||c    Otherwise Nop    End  Say gl '= 0'   d=b**2-4*a*c  If d<0 Then Do    dd=sqrt(-d)    r=-b/(2*a)    i=dd/(2*a)    x1=r'+'i'i'    x2=r'-'i'i'    End  Else Do    dd=sqrt(d)    x1=(-b+dd)/(2*a)    x2=(-b-dd)/(2*a)    End  Say 'x1='||x1  Say 'x2='||x2  Exitsqrt:/* REXX **************************************************************** EXEC to calculate the square root of x with high precision**********************************************************************/  Parse Arg x  prec=digits()  prec1=2*prec  eps=10**(-prec1)  k = 1  Numeric Digits prec1  r0= x  r = 1  Do i=1 By 1 Until r=r0 | (abs(r*r-x)<eps)    r0 = r    r  = (r + x/r) / 2    k  = min(prec1,2*k)    Numeric Digits (k + 5)    End  Numeric Digits prec  Return (r+0)exit: Say arg(1)
Output:
Version 1:
a = 1
b = -1
c = 0

root1 = 1
root2 = 0

Version 2:
1*x**2-1.0000000001*x+1.e-9 = 0
x1=0.9999999991000000000025
x2=0.0000000009999999999975


## Ring

 x1 = 0x2 = 0quadratic(3, 4, 4/3.0)  # [-2/3]see "x1 = " + x1 + " x2 = " + x2 + nlquadratic(3, 2, -1)      # [1/3, -1]see "x1 = " + x1 + " x2 = " + x2 + nlquadratic(-2,  7, 15)    # [-3/2, 5]see "x1 = " + x1 + " x2 = " + x2 + nlquadratic(1, -2,  1)     # [1]see "x1 = " + x1 + " x2 = " + x2 + nl func quadratic a, b, c     sqrtDiscriminant = sqrt(pow(b,2) - 4*a*c)     x1 = (-b + sqrtDiscriminant) / (2.0*a)     x2 = (-b - sqrtDiscriminant) / (2.0*a)     return [x1, x2]

## Ruby

Works with: Ruby version 1.9.3+

The CMath#sqrt method will return a Complex instance if necessary.

require 'cmath' def quadratic(a, b, c)  sqrt_discriminant = CMath.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)   # [-3/2, 5]p quadratic(1, -2,  1)    # [1]p quadratic(1,  3,  3)    # [(-3 + sqrt(3)i)/2), (-3 - sqrt(3)i)/2)]
Output:
[-0.6666666666666666, -0.6666666666666666]
[0.3333333333333333, -1.0]
[(-0.3333333333333333+0.47140452079103173i), (-0.3333333333333333-0.47140452079103173i)]
[(0.0+1.0i), (0.0-1.0i)]
[999999.999999, 1.00000761449337e-06]
[-1.5, 5.0]
[1.0, 1.0]
[(-1.5+0.8660254037844386i), (-1.5-0.8660254037844386i)]


print "FOR 1,2,3 => ";quad$(1,2,3)print "FOR 4,5,6 => ";quad$(4,5,6) 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 ifEND FUNCTION FOR 1,2,3 => -1 +i1.41421356 , -1 -i1.41421356 FOR 4,5,6 => -0.625 +i1.05326872 , -0.625 -i1.05326872 ## Scala Using Complex class from task Arithmetic/Complex. import ArithmeticComplex._object QuadraticRoots { def solve(a:Double, b:Double, c:Double)={ val d = b*b-4.0*a*c val aa = a+a if (d < 0.0) { // complex roots val re= -b/aa; val im = math.sqrt(-d)/aa; (Complex(re, im), Complex(re, -im)) } else { // real roots val re=if (b < 0.0) (-b+math.sqrt(d))/aa else (-b -math.sqrt(d))/aa (re, (c/(a*re))) } }} Usage: val equations=Array( (1.0, 22.0, -1323.0), // two distinct real roots (6.0, -23.0, 20.0), // with a != 1.0 (1.0, -1.0e9, 1.0), // with one root near zero (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); equations.foreach{v => val (a,b,c)=v println("a=%g b=%g c=%g".format(a,b,c)) val roots=solve(a, b, c) println("x1="+roots._1) if(roots._1 != roots._2) println("x2="+roots._2) println} Output: a=1.00000 b=22.0000 c=-1323.00 x1=-49.0 x2=27.0 a=6.00000 b=-23.0000 c=20.0000 x1=2.5 x2=1.3333333333333333 a=1.00000 b=-1.00000e+09 c=1.00000 x1=1.0E9 x2=1.0E-9 a=1.00000 b=2.00000 c=1.00000 x1=-1.0 a=1.00000 b=0.00000 c=1.00000 x1=-0.0 + 1.0i x2=-0.0 + -1.0i a=1.00000 b=1.00000 c=1.00000 x1=-0.5 + 0.8660254037844386i x2=-0.5 + -0.8660254037844386i ## 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) ## Seed7 Translation of: Ada $ include "seed7_05.s7i";  include "float.s7i";  include "math.s7i"; const type: roots is new struct    var float: x1 is 0.0;    var float: x2 is 0.0;  end struct; const func roots: solve (in float: a, in float: b, in float: c) is func  result    var roots: solution is roots.value;  local    var float: sd is 0.0;    var float: x is 0.0;  begin    sd := sqrt(b**2 - 4.0 * a * c);    if b < 0.0 then      x := (-b + sd) / 2.0 * a;      solution.x1 := x;      solution.x2 := c / (a * x);    else      x := (-b - sd) / 2.0 * a;      solution.x1 := c / (a * x);      solution.x2 := x;    end if;  end func; const proc: main is func  local    var roots: r is roots.value;  begin    r := solve(1.0, -10.0E5, 1.0);    writeln("X1 = " <& r.x1 digits 6 <& " X2 = " <& r.x2 digits 6);   end func;
Output:
X1 = 1000000.000000 X2 = 0.000001


## Sidef

var sets = [            [1,    2,  1],            [1,    2,  3],            [1,   -2,  1],            [1,    0, -4],            [1, -1e6,  1],           ] func quadroots(a, b, c) {    var root = sqrt(b**2 - 4*a*c)     [(-b + root) / (2 * a),     (-b - root) / (2 * a)]} sets.each { |coefficients|    say ("Roots for #{coefficients}",        "=> (#{quadroots(coefficients...).join(', ')})")}
Output:
Roots for [1, 2, 1]=> (-1, -1)
Roots for [1, 2, 3]=> (-1+1.41421356237309504880168872420969807856967187538i, -1-1.41421356237309504880168872420969807856967187538i)
Roots for [1, -2, 1]=> (1, 1)
Roots for [1, 0, -4]=> (2, -2)
Roots for [1, -1000000, 1]=> (999999.999998999999999998999999999997999999999995, 0.00000100000000000100000000000200000000000500000000002)


## Stata

mata: polyroots((-2,0,1))                 1             2    +-----------------------------+  1 |   1.41421356   -1.41421356  |    +-----------------------------+ : polyroots((2,0,1))                  1              2    +-------------------------------+  1 |  -1.41421356i    1.41421356i  |    +-------------------------------+

## Tcl

Library: Tcllib (Package: math::complexnumbers)
package require math::complexnumbersnamespace 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"    }} # examples on this pagereport_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}# examples from http://en.wikipedia.org/wiki/Quadratic_equationreport_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)}
Output:
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

## zkl

zkl doesn't have a complex number package.

Translation of: Elixir
fcn quadratic(a,b,c){ b=b.toFloat();   println("Roots of a quadratic function %s, %s, %s".fmt(a,b,c));   d,a2:=(b*b - 4*a*c), a+a;   if(d>0){      sd:=d.sqrt();      println("  the real roots are %s and %s".fmt((-b + sd)/a2,(-b - sd)/a2));   }   else if(d==0) println("  the single root is ",-b/a2);   else{      sd:=(-d).sqrt();      println("  the complex roots are %s and \U00B1;%si".fmt(-b/a2,sd/a2));   }}
foreach a,b,c in (T( T(1,-2,1), T(1,-3,2), T(1,0,1), T(1,-1.0e10,1), T(1,2,3), T(2,-1,-6)) ){   quadratic(a,b,c)}
Output:
Roots of a quadratic function 1, -2, 1
the single root is 1
Roots of a quadratic function 1, -3, 2
the real roots are 2 and 1
Roots of a quadratic function 1, 0, 1
the complex roots are 0 and ±1i
Roots of a quadratic function 1, -1e+10, 1
the real roots are 1e+10 and 0
Roots of a quadratic function 1, 2, 3
the complex roots are -1 and ±1.41421i
Roots of a quadratic function 2, -1, -6
the real roots are 2 and -1.5