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Program to find the Roots of Quadratic equation
  • Difficulty Level : Basic
  • Last Updated : 08 Oct, 2020

Given a quadratic equation in the form ax2 + bx + c, find roots of it. 
 

Examples :

Input  :  a = 1, b = -2, c = 1
Output :  Roots are real and same
          1

Input  :  a = 1, b = 7, c = 12
Output :  Roots are real and different
          -3, -4

Input  :  a = 1, b = 1, c = 1
Output :  Roots are complex 
          -0.5 + i1.73205
          -0.5 - i1.73205  

Below is direct formula for finding roots of quadratic equation.
 

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}



There are following important cases. 

If b*b < 4*a*c, then roots are complex
(not real).
For example roots of x2 + x + 1, roots are
-0.5 + i1.73205 and -0.5 - i1.73205

If b*b == 4*a*c, then roots are real 
and both roots are same.
For example, roots of x2 - 2x + 1 are 1 and 1

If b*b > 4*a*c, then roots are real 
and different.
For example, roots of x2 - 7x - 12 are 3 and 4

Below is the implementation of the above formula.

C




/* C program to find roots of a quadratic equation */
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
 
// Prints roots of quadratic equation ax*2 + bx + x
void findRoots(int a, int b, int c)
{
    // If a is 0, then equation is not quadratic, but
    // linear
    if (a == 0) {
        printf("Invalid");
        return;
    }
 
    int d = b * b - 4 * a * c;
    double sqrt_val = sqrt(abs(d));
 
    if (d > 0) {
        printf("Roots are real and different \n");
        printf("%f\n%f", (double)(-b + sqrt_val) / (2 * a),
               (double)(-b - sqrt_val) / (2 * a));
    }
    else if (d == 0) {
        printf("Roots are real and same \n");
        printf("%f", -(double)b / (2 * a));
    }
    else // d < 0
    {
        printf("Roots are complex \n");
        printf("%f + i%f\n%f - i%f", -(double)b / (2 * a),
               sqrt_val, -(double)b / (2 * a), sqrt_val);
    }
}
 
// Driver code
int main()
{
    int a = 1, b = -7, c = 12;
   
    // Function call
    findRoots(a, b, c);
    return 0;
}


C++




/* C++ program to find roots of a quadratic equation */
#include <bits/stdc++.h>
using namespace std;
 
// Prints roots of quadratic equation ax*2 + bx + x
void findRoots(int a, int b, int c)
{
    // If a is 0, then equation is not quadratic, but
    // linear
    if (a == 0) {
        cout << "Invalid";
        return;
    }
 
    int d = b * b - 4 * a * c;
    double sqrt_val = sqrt(abs(d));
 
    if (d > 0) {
        cout << "Roots are real and different \n";
        cout << (double)(-b + sqrt_val) / (2 * a) << "\n"
             << (double)(-b - sqrt_val) / (2 * a);
    }
    else if (d == 0) {
        cout << "Roots are real and same \n";
        cout << -(double)b / (2 * a);
    }
    else // d < 0
    {
        cout << "Roots are complex \n";
        cout << -(double)b / (2 * a) << " + i" << sqrt_val
             << "\n"
             << -(double)b / (2 * a) << " - i" << sqrt_val;
    }
}
 
// Driver code
int main()
{
    int a = 1, b = -7, c = 12;
   
    // Function call
    findRoots(a, b, c);
    return 0;
}


Java




// Java program to find roots
// of a quadratic equation
 
import java.io.*;
import static java.lang.Math.*;
class Quadratic {
 
    // Prints roots of quadratic
    // equation ax * 2 + bx + x
    static void findRoots(int a, int b, int c)
    {
        // If a is 0, then equation is not
        // quadratic, but linear
 
        if (a == 0) {
            System.out.println("Invalid");
            return;
        }
 
        int d = b * b - 4 * a * c;
        double sqrt_val = sqrt(abs(d));
 
        if (d > 0) {
            System.out.println(
                "Roots are real and different \n");
 
            System.out.println(
                (double)(-b + sqrt_val) / (2 * a) + "\n"
                + (double)(-b - sqrt_val) / (2 * a));
        }
        else if (d == 0) {
            System.out.println(
                "Roots are real and same \n");
 
            System.out.println(-(double)b / (2 * a) + "\n"
                               + -(double)b / (2 * a));
        }
        else // d < 0
        {
            System.out.println("Roots are complex \n");
 
            System.out.println(-(double)b / (2 * a) + " + i"
                               + sqrt_val + "\n"
                               + -(double)b / (2 * a)
                               + " - i" + sqrt_val);
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int a = 1, b = -7, c = 12;
       
        // Function call
        findRoots(a, b, c);
    }
}
 
// This code is contributed by Sumit Kumar.


Python3




# Python program to find roots
# of a quadratic equation
import math
 
# Prints roots of quadratic equation
# ax*2 + bx + x
 
 
def findRoots(a, b, c):
 
    # If a is 0, then equation is
    # not quadratic, but linear
    if a == 0:
        print("Invalid")
        return -1
    d = b * b - 4 * a * c
    sqrt_val = math.sqrt(abs(d))
 
    if d > 0:
        print("Roots are real and different ")
        print((-b + sqrt_val)/(2 * a))
        print((-b - sqrt_val)/(2 * a))
    elif d == 0:
        print("Roots are real and same")
        print(-b / (2*a))
    else# d<0
        print("Roots are complex")
        print(- b / (2*a), " + i", sqrt_val)
        print(- b / (2*a), " - i", sqrt_val)
 
 
# Driver Program
a = 1
b = -7
c = 12
 
# Function call
findRoots(a, b, c)
 
# This code is contributed by Sharad Bhardwaj.


C#




// C# program to find roots
// of a quadratic equation
using System;
 
class Quadratic {
 
    // Prints roots of quadratic
    // equation ax * 2 + bx + x
    void findRoots(int a, int b, int c)
    {
 
        // If a is 0, then equation is
        // not quadratic, but linear
 
        if (a == 0) {
            Console.Write("Invalid");
            return;
        }
 
        int d = b * b - 4 * a * c;
        double sqrt_val = Math.Abs(d);
 
        if (d > 0) {
            Console.Write(
                "Roots are real and different \n");
 
            Console.Write(
                (double)(-b + sqrt_val) / (2 * a) + "\n"
                + (double)(-b - sqrt_val) / (2 * a));
        }
 
        // d < 0
        else {
            Console.Write("Roots are complex \n");
 
            Console.Write(-(double)b / (2 * a) + " + i"
                          + sqrt_val + "\n"
                          + -(double)b / (2 * a) + " - i"
                          + sqrt_val);
        }
    }
 
    // Driver code
    public static void Main()
    {
        Quadratic obj = new Quadratic();
        int a = 1, b = -7, c = 12;
       
        // Function call
        obj.findRoots(a, b, c);
    }
}
 
// This code is contributed by nitin mittal.


PHP




<?php
// PHP program to find roots
// of a quadratic equation
 
// Prints roots of quadratic
// equation ax*2 + bx + x
function findRoots($a, $b, $c)
{
    // If a is 0, then equation is
    // not quadratic, but linear
    if ($a == 0)
    {
        echo "Invalid";
        return;
    }
 
    $d = $b * $b - 4 * $a * $c;
    $sqrt_val = sqrt(abs($d));
 
    if ($d > 0)
    {
        echo "Roots are real and ".
                    "different \n";
        echo (-$b + $sqrt_val) / (2 * $a) , "\n",
             (-$b - $sqrt_val) / (2 * $a);
    }
    else if ($d == 0)
    {
        echo "Roots are real and same \n";
        echo -$b / (2 * $a);
    }
     
    // d < 0
    else
    {
        echo "Roots are complex \n";
        echo -$b / (2 * $a) , " + i" ,
              $sqrt_val, "\n" , -$b / (2 * $a),
                             " - i", $sqrt_val;
    }
}
 
// Driver code
$a = 1; $b = -7 ;$c = 12;
 
// Function call
findRoots($a, $b, $c);
 
// This code is contributed
// by nitin mittal.
?>


Output

Roots are real and different 
4.000000
3.000000
This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
 

 

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