How would I find the imaginary roots given a quadratic? - php

Question

This code finds the two roots of a quadratic equation in the form ax^2+bx+c
Is my solution good time/space-complexity wise, and how I would go about allowing users to see the imaginary roots if the quadratic has any?
public function factor($a=0, $b=0, $c=0) {
$positive_solution = (-$b + sqrt($b**2-4*$a*$c))/2*$a;
$negative_solution = (-$b - sqrt($b**2-4*$a*$c))/2*$a;
if($b**2-4*$a*$c < 0) {
return "Solutions are imaginary";
}
$factor_one = $positive_solution * -1;
$factor_two = $negative_solution * -1;
$factor_one > 0 ? $factor_one = "+ " . $factor_one : $factor_one = "- " . $factor_one;
$factor_two > 0 ? $factor_two = "+ " . $factor_two : $factor_two = "- " . $factor_two;
return "Your roots are located at (0, " . $positive_solution . ")(0, " . $negative_solution . "),
Thus, the problem can be factored into (x " . $factor_one . ")(x " . $factor_two . ")";
}

Answer 1

the if($b**2-4*$a*$c < 0) is too late
You already used the sqrt so domain error might be thrown move the if to start
for imaginary part you just use negation or abs value
simply sqrt(-($b**2-4*$a*$c)) or sqrt(abs($b**2-4*$a*$c)) not sure if php uses abs or fabs for floats or if you even use floats... (havent code in php for ages)
I would combine this to something like this (just pseudo code as the var $ makes me dizzy :) ):
q = b*b-4*a*c
d = sqrt(abs(q))
if (q<0)
{
a0 = -b/(2*a);
b0 = +d/(2*a);
a1 = -b/(2*a);
b1 = -d/(2*a);
}
else
{
a0 = (-b + d)/(2*a);
b0 = 0
a1 = (-b - d)/(2*a);
b1 = 0
}
Where a0+i*b0 and a1+i*b1 are the 2 solutions where i=sqrt(-1)