JS 一元四次方程计算器

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title></title>
<body>

<div align="center">例如输入:a=3, b=6, c=-123, d=-126 e=1080</div>
<form name="form1" action="">
	ax<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e = 0
	<table border="0" align="center" colspan="4" cellspacing="0" cellpadding="0">
		<tbody>
			<tr>
				<td><input type="text" name="aIn" size="5" placeholder="a">x<sup>4</sup> +</td>
				<td><input type="text" name="bIn" size="5" placeholder="b">x³ +</td>
				<td><input type="text" name="cIn" size="5" placeholder="c">x² +</td>
				<td><input type="text" name="dIn" size="5" placeholder="d">x +</td>
				<td><input type="text" name="eIn" size="5" placeholder="e">e</td>
			</tr>
		</tbody>
	</table>
	
	<center>
		<input type="button" name="button1" value="计算" onclick="quad4solve(this.form)">
	</center>
	
	<table border="0" class="innerc" cellspacing="0" cellpadding="3">
		<tbody>
		<tr align="center">
			<td class="inner" colspan="4"><b>结果:</b></td>
		</tr>
		<tr>
			<td>x<sub>1</sub>: 
				<input type="text" name="x1Re" size="25"> + &nbsp;
				<input type="text" name="x1Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>2</sub>: 
				<input type="text" name="x2Re" size="25"> + &nbsp;
				<input type="text" name="x2Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>3</sub>: 
				<input type="text" name="x3Re" size="25"> + &nbsp;
				<input type="text" name="x3Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>4</sub>: 
				<input type="text" name="x4Re" size="25"> + &nbsp;
				<input type="text" name="x4Im" size="25"> i 
			</td>
		</tr>
		</tbody>
	</table>
</form>

<script type="text/javascript">
function quad4solve(dataForm)
{
	var a = parseFloat(dataForm.aIn.value);
	var b = parseFloat(dataForm.bIn.value);
	var c = parseFloat(dataForm.cIn.value);
	var d = parseFloat(dataForm.dIn.value);
	var e = parseFloat(dataForm.eIn.value);
	if (a == 0)
	{
		alert("The coefficient of the power four of x is 0. Please use the utility for a third degree quadratic.");
		return;
	} 
	if (e == 0)
	{
		alert("One root is 0. Now divide through by x and use the utility for a third degree quadratic to solve the resulting equation for the other three roots.");
		return;
	} 
	if (a != 1) 
	{
		b /= a;
		c /= a;
		d /= a;
		e /= a;
	}

	//立方解算器系数
	var cb, cc, cd;  
	var discrim, q, r, RRe, RIm, DRe, DIm, dum1, ERe, EIm, s, t, term1, r13, sqR, y1, z1Re, z1Im, z2Re;
	cb = -c;
	cc = -4.0*e + d*b;
	cd = -(b*b*e + d*d) + 4.0*c*e;
	if (cd == 0)  
	{
		alert("cd = 0.");
	}
	q = (3.0*cc - (cb*cb))/9.0;
	r = -(27.0*cd) + cb*(9.0*cc - 2.0*(cb*cb));
	r /= 54.0;
	discrim = q*q*q + r*r;
	term1 = (cb/3.0);
	if (discrim > 0) 
	{ 
		// 1 实数, 2 复数
		s = r + Math.sqrt(discrim);
		s = ((s < 0) ? -Math.pow(-s, (1.0/3.0)) : Math.pow(s, (1.0/3.0)));
		t = r - Math.sqrt(discrim);
		t = ((t < 0) ? -Math.pow(-t, (1.0/3.0)) : Math.pow(t, (1.0/3.0)));
		y1 = -term1 + s + t;
		} 
		else 
		{
			if (discrim == 0) 
			{
				r13 = ((r < 0) ? -Math.pow(-r,(1.0/3.0)) : Math.pow(r,(1.0/3.0)));
				y1 = -term1 + 2.0*r13;
			} 
			else 
			{                             	
				q = -q;
				dum1 = q*q*q;
				dum1 = Math.acos(r/Math.sqrt(dum1));
				r13 = 2.0*Math.sqrt(q);
				y1 = -term1 + r13*Math.cos(dum1/3.0);
			}
		} 
		// 确定了y1 分解立方的实根
		term1 = b/4.0;
		sqR = -c + term1*b + y1;  
		RRe = RIm = DRe = DIm = ERe = EIm = z1Re = z1Im = z2Re = 0;
		if (sqR >= 0) 
		{
			if (sqR == 0) 
			{
				dum1 = -(4.0*e) + y1*y1;
				if (dum1 < 0){
					z1Im = 2.0*Math.sqrt(-dum1);
				} else {
	  				z1Re = 2.0*Math.sqrt(dum1);
					z2Re = -z1Re;
				}
			} 
			else 
			{                       
			RRe = Math.sqrt(sqR);
			z1Re = -(8.0*d + b*b*b)/4.0 + b*c;
			z1Re /= RRe;
			z2Re = -z1Re;
		 } 
	} 
	else 
	{                           
		RIm = Math.sqrt(-sqR);
		z1Im = -(8.0*d + b*b*b)/4.0 + b*c;
		z1Im /= RIm;
		z1Im = -z1Im;
	} 
	z1Re += -(2.0*c + sqR) + 3.0*b*term1;
	z2Re += -(2.0*c + sqR) + 3.0*b*term1;

	//At this point, z1 and z2 should be the terms under the square root for D and E
	if (z1Im == 0)
	{               // Both z1 and z2 real
		if (z1Re >= 0)
		{
			DRe = Math.sqrt(z1Re);
		}
		else
		{
			DIm = Math.sqrt(-z1Re);
		}
		if (z2Re >= 0)
		{
			ERe = Math.sqrt(z2Re);
		}
		else
		{
			EIm = Math.sqrt(-z2Re);
		}
	}
	else 
	{                      
		r = Math.sqrt(z1Re*z1Re + z1Im*z1Im); 
		r = Math.sqrt(r);
		dum1 = Math.atan2(z1Im, z1Re); 
		dum1 /= 2; //Divide this angle by 2
		ERe = DRe = r*Math.cos(dum1); 
		DIm = r*Math.sin(dum1);
		EIm = -DIm;
	} 
	dataForm.x1Re.value = -term1 + (RRe + DRe)/2;
	dataForm.x1Im.value = (RIm + DIm)/2;
	dataForm.x2Re.value = -(term1 + DRe/2) + RRe/2;
	dataForm.x2Im.value = (-DIm + RIm)/2;
	dataForm.x3Re.value = -(term1 + RRe/2) + ERe/2;
	dataForm.x3Im.value = (-RIm + EIm)/2;
	dataForm.x4Re.value = -(term1 + (RRe + ERe)/2);
	dataForm.x4Im.value = -(RIm + EIm)/2;
	return;
}
</script>
</body>
</html>

1,886 thoughts on “JS 一元四次方程计算器”

  1. Great beat ! I would like to apprentice while you amend your site, how
    can i subscribe for a blog website? The account
    helped me a acceptable deal. I had been tiny bit acquainted of
    this your broadcast provided bright clear concept

  2. Unquestionably believe that which you stated. Your favourite
    justification appeared to be on the web the simplest factor
    to have in mind of. I say to you, I definitely get annoyed whilst other people think about issues that they plainly do not recognise about.
    You controlled to hit the nail upon the highest and defined out the entire thing without having side-effects ,
    people could take a signal. Will likely be again to get more.

    Thank you

发表回复

您的电子邮箱地址不会被公开。 必填项已用*标注

Back To Top

鄂ICP备17008157号-1