Quadratics

[Bulbe]

Hey everyone,

i've been doing a work in math and it should be about solving quadratics in php. Well, my dictonary says that quadratics mean all this stuff, not only ax²+bx+c. So my question is, do you think it would be possible to make this in php ? You would enter the quadratic and it would show its roots

 

Best regard, Bulb

[AndyB]

Trivial exercise, surely.

 

http://en.wikipedia.org/wiki/Quadratic_equation

[GingerRobot]

Yeah, very easy - just use the quadratic formula. Check the discriminant to work out if you're going to get any real answers, then plug the values of a,b and c into the formula. You could set it up to give you non-real answers too.

[Bulbe]

yes, but imagine somone wants to solve e.g. 2x²+32-2x=3x+2x-32-x²

[GingerRobot]

It wouldn't be that difficult to have it rearranged and solved.

[GingerRobot]

Was bored:

 

<?php
$eqn = '-2x^2+32-2x=3x+2x-32-x^2';
function rearrange($eqn){
    $sides = explode('=',$eqn);
    $coefficients = array();
    foreach($sides as $k => $v){
        preg_match_all('%(\-|\+)?[0-9x^]+%s',$v,$terms);

        foreach($terms[0] as $v2){
            //work out which term is being used
            if(strpos($v2, 'x^2') !== false){
                $key = 0;
            }elseif(strpos($v2, 'x') !== false){
                $key = 1;
            }else{
                $key = 2;
            }
		//work out the coefficient
		preg_match('%^(\-|\+)?[0-9]+%s',$v2,$matches);
		if(empty($matches)){
		    if($v2[0] == '-'){
		        $num = -1;
		    }else{
		        $num = 1;
		    }
		}else{
		    $num = $matches[0];
		}
		@$coefficients[$k][$key] += $num;//+= because terms may not be collected
        }
    }
    for($x=0;$x<=2;$x++){
        //collect terms. Surpress error since coeffiecent may not be set
        @$rearranged[$x] = $coefficients[0][$x] - $coefficients[1][$x];
    }
    return $rearranged;
    

}
function solve($coefficients){
    list($a,$b,$c) = $coefficients;
    $discriminant = pow($b,2) - 4*$a*$c;
    if($discriminant > 0){
        $roots[0] = (-$b + sqrt($discriminant))/(2*$a);
        $roots[1] = (-$b - sqrt($discriminant))/(2*$a);        
    }elseif($discriminant == 0){
        $roots[0] = -$b/(2*$a);
    }else{
        $real = -$b/(2*$a);
        $imaginary = sqrt(-$discriminant)/(2*$a);
        $roots[0] = $real.' + '.$imaginary.'i';
        $roots[1] = $real.' - '.$imaginary.'i';
    }
    return $roots;
}
print_r(solve(rearrange($eqn)));
?>

 

Will work as long as there's nothing unexpected in the input.

[cooldude832]

without PEAR php gets very confused on imaginary numbers so any solution should verify that the radical isn't negative pre attempting to calculate it as php gets confused trying to do sqrt(-1n) where n is any thing above 0.

[GingerRobot]

Was that in response to my code? It handes the imaginary numbers fine.

[cooldude832]

Was that in response to my code? It handes the imaginary numbers fine.

 

yes cause you verify the descrimenent I was saying in general php has no ability to handle complex answers (although it should)

[Barand]

There's your next project. A PHP extension to handle complex numbers.