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How to find the roots of a quadratic equation in PHP? _PHP Tutorial

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WBOYOriginal
2016-07-13 10:33:513091browse

Today someone asked a math question, 4x+1/x=2, what is x? Convert the equation, 4x2 + 1 = 2x, and then 4x2 - 2x + 1 =0, which is actually a quadratic equation problem. I haven’t done these things in a long time, and I’m a high school teacher in mathematics. Fortunately, I can write a program. Let's use the program to find the roots of this equation.

<? 
//ax*x bx c=0;  一元二次方程一般形式  
  
//系数设定  
$a = 2;  
$b = 3;  
$c = 0;  
echo '一元二次方程为';
echo $a.'x2'.'+'.$b.'x'.'+'.$c;
  
//求根的函数 
function get_root($a,$b,$c) 
{ 
	//放根的数组  
	$x=0;  
	$x=array();  
	if($a==0) 
	{ 
  		if($b==0) 
     	if($c==0) 
        { 
        	$x[0]=0; 
        	$x[1]="no root"; 
        } 
      	else 
        { 
        	$x[0]="no root"; 
        	$x[1]="no root"; 
        } 
  		else if($b!=0) 
     	{ 
      		$x[0]=(0-$c)/$b; 
      		$x[1]="no root"; 
     	} 
	} 
	else 
	{ 
		//标志  
  		$flg=$b*$b-4*$a*$c;  
		//△ >0 两个不同的根  
  		if($flg >0)  
    	{  
     		$x[0]=((0-$b)+sqrt($flg))/2/$a;  
     		$x[1]=((0-$b)-sqrt($flg))/2/$a;  
    	}  
  		else if($flg==0)//△=0 两个相同的根  
    	{  
    		$x[0]=(0-$b)/2/$a;  
    		$x[1]=(0-$b)/2/$a;  
    	}  
  		else  // 无根  
    	{  
    		$x[0]="no root";  
    		$x[1]="no root";  
    	}  
  	} 
  	return $x; 
} 
  
//验证代码 参数为顶部设置的a b c  的值,可自行修改测试 
$root=array(); 
$root=get_root($a,$b,$c); 
echo " <pre class="brush:php;toolbar:false">求得根: <br>"; 
print_r($root); 
echo " <pre class="brush:php;toolbar:false">"; 
  
?>

The result of running the program is:

一元二次方程为2x2+3x+0
求得根: 
Array
(
    [0] => 0
    [1] => -1.5
)

Back to the original question, after calculation by the program, the result is:

一元二次方程为4x2+-2x+1
求得根: 
Array
(
    [0] => no root
    [1] => no root
)

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/752401.htmlTechArticleToday someone asked a math problem, 4x+1/x=2, what is x. Convert the equation, 4x2 + 1 = 2x, and then 4x2 - 2x + 1 =0, which is actually a quadratic equation problem. Long time no see...
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