上次做数学题,解方程解的难受,于是乎,在参考别人源码的过程中,写出了自己的计算器

<html>
<head>
<meta http-equiv="Content-Type" content="text/html" charset="utf-8">
<title>在线一元二次方程式计算器</title>
</head>
<body>
<form name="fquad">
<p align="center">解二次方程式计算<br>
</p>
<table align="center">
<tbody>
<tr>
<td bgcolor="#990000">
<h2><font color="#ffffff"><input size="4" name="fa" type="text"> x<sup>2</sup>+ <input size="4" name="fb" type="text"> x + <input size="4" name="fc" type="text"> = 0 <input onclick="checkQuad()" type="button" value="解题"> <input type="reset" value="重置"> </font></h2>
<p align="center"><font color="#ffffff" face="Arial"><b>一元二次方程的解法</b></font></p>
</td>
</tr>
<tr>
<td bgcolor="#990000">
<h2><font color="#ffffff">x<sub><a style="text-decoration: none" ><font color="#ffffff">1</font></a></sub>=<input size="45" name="x1" type="text"> <br>
x<sub>2</sub>=<input size="45" name="x2" type="text"> </font></h2>
</td>
</tr>
<tr>
</tr>
</tbody>
</table>
</form>
<p align="center">Made by CRoot</p>
<script language="JavaScript">
<!-- 
var rootparti;
var rootpart;
var det;
var rootparti1;
var rootparti2;
var a;
var b;
var c;
var x1;
 var x2;
 var i = "i";
 function checkQuad() {
 var a = document.fquad.fa.value;
 var b = document.fquad.fb.value;
 var c = document.fquad.fc.value;
 if (a == 0 && c != 0) {
 x1 = -c / b;
 x2 = "Not a quadratic equation, but here is your answer for x";
 document.fquad.x1.value=x1;
 document.fquad.x2.value=x2;
 }
 else if (a == "" && c != 0) {
 x1 = -c / b;
 x2 = "Not a quadratic equation";
 document.fquad.x1.value=x1;
 document.fquad.x2.value=x2;
 }
 else {
 quad();
    }
 }
 function quad() {
 var a = document.fquad.fa.value;
 var b = document.fquad.fb.value;
 var c = document.fquad.fc.value;
 det = Math.pow(b,2) - 4 * a * c;
 rootpart = Math.sqrt(det) / (2 * a);
 rootparti = (Math.sqrt(-det) / (2 * a)) + i;
 if (parseFloat(rootparti) < 0) {
 rootparti1 = rootparti;
 rootparti2 = (-1 * parseFloat(rootparti)) + i;
 }
 else {
 rootparti1 = (-1 * parseFloat(rootparti)) + i;
 rootparti2 = rootparti;
 }
 if (rootparti1 == "1i") {
 rootparti1 = i;
 rootparti2 = "-i";
 }
 else if (rootparti1 == "-1i") {
 rootparti1 = "-i";
 rootparti2 = i;
  }
  if (det == 0) {
  x1 = x2 = -b / (2 * a);
  }
  else if (det > 0) {
  x1 = (-b + Math.sqrt(det)) / (2 * a);
  x2 = (-b - Math.sqrt(det)) / (2 * a);
  }
  else if ((-b / (2 * a)) == 0) {
  x1 = rootparti1;
  x2 = rootparti2;
  }
  else {
  x1 = (-b / (2 * a) + " + " + rootparti1);
 x2 = (-b / (2 * a) + " + " + rootparti2);
 }
 document.fquad.x1.value=x1;
 document.fquad.x2.value=x2;
 }
 // will solve for complex numbers
 
 //   -->
 </script>
 </body>
 </html>

 

解二次方程式计算

x2+ x + = 0

一元二次方程的解法

x1=

x2=

Made by CRoot

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