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Program to find the Roots of a Quadratic Equation

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  • Difficulty Level : Easy
  • Last Updated : 13 Sep, 2022
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Given a quadratic equation in the form ax2 + bx + c, (Only the values of a, b and c are provided) the task is to find the roots of the equation.

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  

Roots of Quadratic Equation using Sridharacharya Formula:

The roots could be found using the below formula (It is known as the formula of Sridharacharya)

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

The values of the roots depends on the term (b2 – 4ac) which is known as the discriminant (D)

If D > 0:
        => This occurs when b2 > 4ac.
        => The roots are real and unequal.
        => The roots are {-b + √(b2 – 4ac)}/2a and {-b – √(b2 – 4ac)}/2a.

If D = 0:
        => This occurs when b2 = 4ac.
        => The roots are real and equal.
        => The roots are (-b/2a).

If D < 0:
        => This occurs when b2 < 4ac.
        => The roots are imaginary and unequal.
        => The discriminant can be written as (-1 * -D).
        => As D is negative, -D will be positive.
        => The roots are {-b ± √(-1*-D)} / 2a = {-b ± i√(-D)} / 2a = {-b ± i√-(b2 – 4ac)}/2a where i = √-1.

Use the following pseudo algorithm to find the roots of the 

Pseudo algorithm:

Start
Read the values of a, b, c
Compute d = b2 – 4ac
If d > 0
    calculate root1 = {-b + √(b2 – 4ac)}/2a
    calculate root2 = {-b – √(b2 – 4ac)}/2a
else If d = 0
    calculate root1 = root2 = (-b/2a)
else
    calculate root1 = {-b + i√-(b2 – 4ac)}/2a
    calculate root2 = {-b + i√-(b2 – 4ac)}/2a
print root1 and root2
End

Below is the implementation of the above formula.

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 / (2 * a) << "\n"
             << -(double)b / (2 * a) << " - i"
             << sqrt_val / (2 * a);
    }
}
 
// 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 <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 / (2 * a), -(double)b / (2 * a),
               sqrt_val / (2 * a));
    }
}
 
// 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 static java.lang.Math.*;
 
import java.io.*;
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 / (2 * a) + "\n"
                               + -(double)b / (2 * a)
                               + " - i"
                               + sqrt_val / (2 * a));
        }
    }
 
    // 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 / (2 * a))
        print(- b / (2*a), " - i", sqrt_val / (2 * a))
 
 
# Driver Program
if __name__ == '__main__':
    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 / (2 * a) + "\n"
                          + -(double)b / (2 * a) + " - i"
                          + sqrt_val / (2 * a));
        }
    }
 
    // 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 / (2 * $a) , "\n" , -$b / (2 * $a),
                             " - i", $sqrt_val / (2 * $a) ;
    }
}
 
// Driver code
$a = 1; $b = -7 ;$c = 12;
 
// Function call
findRoots($a, $b, $c);
 
// This code is contributed
// by nitin mittal.
?>

Javascript




<script>
 
// JavaScript 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) {
            document.write("Invalid");
            return;
        }
  
        let d = b * b - 4 * a * c;
        let sqrt_val = Math.sqrt(Math.abs(d));
  
        if (d > 0) {
            document.write(
                "Roots are real and different \n" + "<br/>");
  
            document.write(
                (-b + sqrt_val) / (2 * a) + "<br/>"
                + (-b - sqrt_val) / (2 * a));
        }
        else if (d == 0) {
            document.write(
                "Roots are real and same \n" + "<br/>");
  
            document.write(-b / (2 * a) + "<br/>"
                               + -b / (2 * a)) ;
        }
        else // d < 0
        {
            document.write("Roots are complex \n");
  
            document.write(-b / (2 * a) + " + i"
                               + sqrt_val / (2 * a)  + "<br/>"
                               + -b / (2 * a)
                               + " - i" + sqrt_val) / (2 * a) ;
        }
    }
 
// Driver Code
 
        let a = 1, b = -7, c = 12;
        
        // Function call
        findRoots(a, b, c);
 
</script>

Output

Roots are real and different 
4.000000
3.000000

Time Complexity: O(log(D)), where D is the discriminant of the given quadratic equation.
Auxiliary Space: O(1)

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This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or have more information about the topic discussed above. 


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