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I prefer the solution involving radicals.

What's cool is that you can use square roots to solve quadratics, third roots to solve cubics, fourth roots to solve quartics, but you can't use just radicals from 5 and up.

Perl 6's built-in roots function comes in handy.

This equation comes right from the wikipedia entry, and uses the names of the variables there.
#!/usr/bin/env perl6

sub cubic(\a,\b,\c,\d) {
    my \Δ0 = b	- 3  a  c;
    # note: special case when Δ0 == 0
    my \Δ1 = 2 * b - 9  a  b  c + 27  a  d;
    my \C = ( ( Δ1 + sqrt( Δ1 - 4  Δ0 + 0i) ) / 2 ).roots(3)[0];
    my \ς = 1.roots(3);  # cubic roots of unity
    return [0,1,2].map: -> \k {
        ( -1 / ( 3  a ) )  ( b + ς[k]  C + Δ0 / ( C  ς[k] ) )
    }
}


my @vals = cubic(1,10,10,-10);

# test
use Test;
plan 3;

my $f = -> \x { x + 10 * x + 10 * x - 10 };

is-approx $f( @vals[0] ), 0, 'first value';
is-approx $f( @vals[1] ), 0, 'second value';
is-approx $f( @vals[2] ), 0, 'third value';

I had a hard time doing this with a code block on perlmonks, so I made a gist instead.

In reply to Re: solve cubic equations by bduggan
in thread solve cubic equations by no_slogan

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