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+ 1

Hello please I have a school course work on c++ please I don't know if anyone can help

The question is Write a program that can solve the quadratic equation we call the almighty formula

2nd Jun 2023, 5:16 PM
GOODNESS ODIDO
10 Answers
+ 7
We're not here to write code for you from scratch, but make some kind of attempt first and we'll be happy to help you improve and debug it
2nd Jun 2023, 5:20 PM
Orin Cook
Orin Cook - avatar
+ 2
Good place to start is knowing what the quadratic equation is and how to solve it. Once you understand that, you can break it down into C++ code following the same logic. If they're asking you to do that, then I assume they've already taught you the stuff to put it together. Post up your code if you run into issues, as Orin said, and we'll help out. Resource for understanding the quadratic equation: https://en.wikipedia.org/wiki/Quadratic_equation
2nd Jun 2023, 9:22 PM
AgentSmith
+ 1
Well sure, most linear equations do have a x-intercept lol; but the equation is also no longer quadratic in that case, so an error isn't uncalled for either. I think either option is a defensible choice.
4th Jun 2023, 1:17 AM
Orin Cook
Orin Cook - avatar
+ 1
Supplementally, if you do decide to handle the linear case, don't forget to handle the constant case, a == b == 0 as well (no solutions for c != 0, infinite solutions for c == 0)
4th Jun 2023, 1:20 AM
Orin Cook
Orin Cook - avatar
3rd Jun 2023, 7:14 AM
Bob_Li
Bob_Li - avatar
0
#include <iostream> #include <cmath> int main() { double a, b, c; std::cout << "Enter the coefficient a: "; std::cin >> a; std::cout << "Enter the coefficient b: "; std::cin >> b; std::cout << "Enter the coefficient c: "; std::cin >> c; double discriminant = b * b - 4 * a * c; if (discriminant > 0) { double root1 = (-b + sqrt(discriminant)) / (2 * a); double root2 = (-b - sqrt(discriminant)) / (2 * a); std::cout << "Root 1: " << root1 << std::endl; std::cout << "Root 2: " << root2 << std::endl; } else if (discriminant == 0) { double root = -b / (2 * a); std::cout << "Root: " << root << std::endl; } else { std::cout << "The quadratic equation has no real roots." << std::endl; } return 0; }
3rd Jun 2023, 10:09 PM
GOODNESS ODIDO
0
This is one of them
3rd Jun 2023, 10:10 PM
GOODNESS ODIDO
0
Seems to be working; only thing I'd really add is throwing an error if a==0
4th Jun 2023, 12:50 AM
Orin Cook
Orin Cook - avatar
0
Orin Cook a = 0 shouldn't throw an error... there is a real number solution for x. Most online quadratic solvers will not work for a = 0, but if you try Wolfram Alpha, https://www.wolframalpha.com/input?i=quadratic+formula+calculator&assumption=%7B%22F%22%2C+%22QuadraticEquation%22%2C+%22c%22%7D+-%3E%221%22&assumption=%7B%22FS%22%7D+-%3E+%7B%7B%22QuadraticEquation%22%2C+%22x%22%7D%7D&assumption=%7B%22F%22%2C+%22QuadraticEquation%22%2C+%22b%22%7D+-%3E%222%22&assumption=%7B%22F%22%2C+%22QuadraticEquation%22%2C+%22a%22%7D+-%3E%220%22 It gives you the proper solution. Maybe solving for x when a = 0 should be added to the code?
4th Jun 2023, 1:04 AM
Bob_Li
Bob_Li - avatar
0
#include <iostream> int main() { double x, y; std::cout << "Enter the value of x: "; std::cin >> x; std::cout << "Enter the value of y: "; std::cin >> y; double sum = x + y; double difference = x - y; double product = x * y; double quotient = x / y; std::cout << "Sum of x and y: " << sum << std::endl; std::cout << "Difference of x and y: " << difference << std::endl; std::cout << "Product of x and y: " << product << std::endl; std::cout << "Quotient of x and y: " << quotient << std::endl; return 0; }
4th Jun 2023, 4:26 PM
GOODNESS ODIDO