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 = b² - 3 × a × c;
# note: special case when Δ0 == 0
my \Δ1 = 2 * b³ - 9 × a × b × c + 27 × a² × d;
my \C = ( ( Δ1 + sqrt( Δ1² - 4 × Δ0³ + 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 { x³ + 10 * x² + 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.