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Program to find the Roots of a Quadratic Equation

Last Updated : 20 Mar, 2024
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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)

[Tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}          [/Tex]

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.

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>

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. ?>


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)

Self Paced Course

Self Paced Course

Using Formula in python:

Approach:

  • Import the math module for square root and other mathematical operations.
  • Declare the coefficients a, b, and c of the quadratic equation.
  • Calculate the discriminant discriminant using the formula b^2 – 4ac.
  • Check if the discriminant is greater than zero, zero, or less than zero.
  • If the discriminant is greater than zero, calculate two real and distinct roots using the formula (-b + sqrt(discriminant)) / (2*a) and (-b – sqrt(discriminant)) / (2*a), and print the roots with the message “Roots are real and distinct”.
  • If the discriminant is equal to zero, calculate one real and same root using the formula -b / (2*a), and print the root with the message “Roots are real and same”.
  • If the discriminant is less than zero, calculate two complex and different roots using the formula (-b / 2*a) +/- (sqrt(-discriminant) / (2*a)), and print the roots with the message “Roots are complex and different”.
Python3

import math a = 1 b = -2 c = 1 # Calculate the discriminant discriminant = b ** 2 - 4 * a * c if discriminant > 0: root1 = (-b + math.sqrt(discriminant)) / (2 * a) root2 = (-b - math.sqrt(discriminant)) / (2 * a) print("Roots are real and distinct") print("Root 1:", root1) print("Root 2:", root2) elif discriminant == 0: root = -b / (2 * a) print("Roots are real and same") print("Root:", root) else: realPart = -b / (2 * a) imaginaryPart = math.sqrt(-discriminant) / (2 * a) print("Roots are complex and different") print("Root 1:", realPart, "+", imaginaryPart, "i") print("Root 2:", realPart, "-", imaginaryPart, "i")


Output

Roots are real and same Root: 1.0

Time Complexity: O(1)
Space Complexity: O(1)

 



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