Quadratic Equation Roots
Write a function to calculate roots of a quadratic equation of the from ax^2+bx+c=0
The formula to find the roots of a quadratic equation is given below:
X+ = (-b + root of (b^2 – 4ac))/2a
X- = (-b – root of (b^2 – 4ac))/2a
Input specification:
Input 1: A double value representing the coefficient value of x^2 in a quadratic equation.
Input 2: A double value representing the coefficient value of x in a quadratic equation.
Input 3: A double value representing the coefficient value of 1 in a quadratic equation.
Output Specification:
Return a double array representing the value of the two roots.
Note: Consider the output root values off to 3 decimal places. Assume all roots are real.
Example 1:
Input 1: 1
Input 2: -2
Input 3: -3
Output: {3.0,-1.0}
Explanation:
On substituting the values of a,b and c in the formula, the roots will be as follows:
x+ = 3.0
x- = -1.0
Example 2:
Input 1: 2
Input 2: 4
Input 3: 1
Output: {-1.707, -0.293}
Solution in Python 3:
#https://codewindow.in
#join our telegram channel
import math
a=int(input())
b=int(input())
c=int(input())
pos = (-b + math.sqrt((b**2) - (4*a*c)))/(2*a)
neg = (-b - math.sqrt((b**2) - (4*a*c)))/(2*a)
print(round(neg,3),round(pos,3))
# Telegram @codewindowSolution in C :
//https://codewindow.in
//join our telegram channel @codewindow
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
void findRoots(int a, int b, int c)
{
if (a == 0) {
printf("Invalid");
return;
}
int d = b * b - 4 * a * c;
double sqrt_val = sqrt(abs(d));
if (d > 0) {
printf("%.2f %.2f", (double)(-b + sqrt_val) / (2 * a), (double)(-b - sqrt_val) / (2 * a));
}
else if (d == 0) {
printf("%.2f", -(double)b / (2 * a));
}
else
{
printf("%.2f + i%.2f %.2f - i%.2f", -(double)b / (2 * a),sqrt_val/(2 * a), -(double)b / (2 * a), sqrt_val/(2 * a));
}
}
int main()
{
int a, b, c;
scanf("%d",&a);
scanf("%d",&b);
scanf("%d",&c);
findRoots(a, b, c);
return 0;
}
Solution in C++ :
//https://codewindow.in
//join our telegram channel @codewindow
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
void findRoots(int a, int b, int c)
{
if (a == 0) {
cout << "Invalid";
return;
}
int d = b * b - 4 * a * c;
double sqrt_val = sqrt(abs(d));
if (d > 0) {
cout << (double)(-b + sqrt_val) / (2 * a) << " " << (double)(-b - sqrt_val) / (2 * a);
}
else if (d == 0) {
cout << -(double)b / (2 * a);
}
else
{
cout << -(double)b / (2 * a) << " + i" << sqrt_val << " " << -(double)b / (2 * a) << " - i" << sqrt_val;
}
}
int main()
{
int a, b, c;
cin >> a;
cin >> b;
cin >> c;
findRoots(a, b, c);
return 0;
}Solution in JAVA :
//https://codewindow.in
//join our telegram channel @codewindow
import java.io.*;
import java.lang.*;
import java.util.*;
class CodeWindow {
static void findRoots(int a, int b, int c)
{
if (a == 0) {
System.out.println("Invalid");
return;
}
int d = b * b - 4 * a * c;
double sqrt_val = Math.sqrt(Math.abs(d));
if (d > 0) {
System.out.println((double)(-b + sqrt_val) / (2 * a) + " " + (double)(-b - sqrt_val) / (2 * a));
}
else if (d == 0) {
System.out.println(-(double)b / (2 * a) + " " + -(double)b / (2 * a));
}
else
{
System.out.println(-(double)b / (2 * a) + " + i" + sqrt_val + " " + -(double)b / (2 * a) + " - i" + sqrt_val);
}
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
findRoots(a,b,c);
}
}Output:
-1
2
3
3.0 -1.02
4
1
-1.707 -0.293