Problem Statement

Write a program to find the roots of a quadratic equation, i.e., solve the equation

a

x

2

b x + c

0

{\displaystyle ax^{2}+bx+c=0}

. Your program must correctly handle non-real roots, but it need not check that

a ≠ 0

{\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

a

1

{\displaystyle a=1}

,

b

−

10

5

{\displaystyle b=-10^{5}}

, and

c

1

{\displaystyle c=1}

. (For double-precision floats, set

b

−

10

9

{\displaystyle b=-10^{9}}

.) Consider the following implementation in Ada: As we can see, the second root has lost all significant figures. The right answer is that X2 is about

10

− 6

{\displaystyle 10^{-6}}

. The naive method is numerically unstable. Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters

q

a c

/

b

{\displaystyle q={\sqrt {ac}}/b}

and

f

1

/

2 +

1 − 4

q

2

/

2

{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}

and the two roots of the quardratic are:

− b

a

f

{\displaystyle {\frac {-b}{a}}f}

and

− c

b f

{\displaystyle {\frac {-c}{bf}}}

Task: do it better. This means that given

a

1

{\displaystyle a=1}

,

b

−

10

9

{\displaystyle b=-10^{9}}

, and

c

1

{\displaystyle c=1}

, both of the roots your program returns should be greater than

10

− 11

{\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

b

−

10

6

{\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.

Let's start with the solution:

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 1
QuadraticRoots(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 numbers
QuadraticRoots(2.0, 2.0, -1.0);
# [ 0.366025, -1.36603 ]