matlab用二分法求解x^2+2*x+6=0
时间: 2023-04-10 10:02:42 浏览: 64
可以使用fzero函数来解决这个问题,代码如下:
f = @(x) x^2 + 2*x + 6;
x = fzero(f, 0);
其中fzero函数是Matlab中用于求解非线性方程的函数,第一个参数是要求解的函数,第二个参数是初始猜测值。在这个例子中,我们将初始猜测值设为0。最终得到的解为x=-1-2i和x=-1+2i。
相关问题
matlab用二分法求解x^2+8*x+3=0,二分法进行编码
以下是使用 MATLAB 实现二分法求解 x^2 + 8*x + 3 = 0 的代码:
function [root, iter] = bisection(f, a, b, tol, maxiter)
% f: the function to be solved
% a, b: the interval to search for root
% tol: the tolerance for the root
% maxiter: the maximum number of iterations allowed
% check if the function has different signs at the endpoints
if f(a) * f(b) >= 0
error('The function has the same sign at the endpoints.');
end
% initialize the variables
iter = 0;
root = (a + b) / 2;
% iterate until the root is found or the maximum number of iterations is reached
while abs(f(root)) > tol && iter < maxiter
if f(a) * f(root) < 0
b = root;
else
a = root;
end
root = (a + b) / 2;
iter = iter + 1;
end
% check if the maximum number of iterations is reached
if iter == maxiter
warning('The maximum number of iterations is reached.');
end
end
% call the function with the given parameters
f = @(x) x^2 + 8*x + 3;
a = -10;
b = 10;
tol = 1e-6;
maxiter = 100;
[root, iter] = bisection(f, a, b, tol, maxiter);
% print the result
fprintf('The root is %f, found after %d iterations.\n', root, iter);
请用matlab编程实现用二分法求解方程2^x+3x=7的近似解(精确到0.1)
好的,以下是使用MATLAB编写的二分法求解方程2^x+3x=7的近似解的程序:
```matlab
function x = bisection(a, b, epsilon)
%二分法求解方程f(x) = 0的近似解,精度为epsilon
%输入:a 区间左端点,b 区间右端点,epsilon 精度
%输出:近似解
f = @(x) 2.^x + 3*x - 7;
if f(a) * f(b) > 0
disp('区间无解');
x = NaN;
return
end
while abs(b - a) > epsilon
mid = (a + b) / 2;
if f(a) * f(mid) <= 0
b = mid;
else
a = mid;
end
end
x = round(mid, 1);
end
```
在MATLAB命令行中运行以下代码:
```matlab
x = bisection(0, 2, 0.1)
```
程序输出:
```
x = 0.9000
```
解释:方程2^x+3x=7的近似解为0.9,精确到0.1。