a) x1 = complex(real_part, imaginary_part) x2 = complex(real_part, -imaginary_part) print(f'方程有两个不同的虚数根:x1={x1:.2f}, x2={x2:.2f}') return x1, x2 # 使用这个类来解方程 a = 1, b = -8, c = 12 equation_solver = Quadrati
The trapezoidal rule uses straight-line segments for approximations, but we can use curves instead. Simpson’s rule fits intervals with quadratic curves. We requi…阅读全文 赞同6 添加评论 分享收藏 Solve Laplace's equation using the Jacobi method Use the Gauss–Seidelmethod ...
foriinrange(0,10):# Loop over i ten times and print out the value of i, followed by a# new line characterprint(i,'\n') 有时,如果代码技术含量高,那么在块注释中使用多个段落是必要的 defquadratic(a,b,c,x):# Calculate the solution to a quadratic equation using the quadratic# formula....
"""Solve quadratic equation via the quadratic formula. A quadratic equation has the following form: ax**2 + bx + c = 0 There always two solutions to a quadratic equation: x_1 & x_2. """ x_1 = (- b+(b**2-4*a*c)**(1/2)) / (2*a) x_2 = (- b-(b**2-4*a*c)...
defquadratic(a,b,c,x):# Calculate the solution to a quadratic equation using the quadratic # formula.# # There are always two solutions to a quadratic equation,x_1 and x_2.x_1=(-b+(b**2-4*a*c)**(1/2))/(2*a)x_2=(-b-(b**2-4*a*c)**(1/2))/(2*a)returnx_1,x...
After all, the intermediate form x2 = -1 of the quadratic equation is the very definition of the imaginary unit. But, wait a minute. Where did the other complex root go? What about complex roots of higher-degree polynomials?For example, a fourth-degree polynomial x4 + 1, which can be...
for i in range(0, 10):# Loop over i ten times and print out the value of i, followed by a# new line characterprint(i, '\n') 有时,如果代码非常技术性,那么必须在块注释中使用多个段落: def quadratic(a, b, c, x):# Calculate the solution to a quadratic equation using the quadratic...
quadraticequation.py 这个程序的名称为 quadratic equation 组合,是二次方程的英文词组。 这个程序用来求解二次方程式: #!/usr/bin/env python3 import math a = int(input("Enter value of a: ")) b = int(input("Enter value of b: "))
For example, instead of calculating the circumference of a circle with 2πr, we can substitute tau and use the simpler equation τr.The use of tau as the circle constant, however, is still under debate. You have the freedom to use either 2π or τ as necessary....
There always two solutions to a quadratic equation: x_1 & x_2. """ x_1 = (- b+(b**2-4*a*c)**(1/2)) / (2*a) x_2 = (- b-(b**2-4*a*c)**(1/2)) / (2*a) return x_1, x_2 1. 2. 3. 4. 5.