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Check if roots of a Quadratic Equation are reciprocal of each other or not

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Given three numbers A, B, C which represents the coefficients(constants) of a quadratic equation Ax^{2} + Bx + C = 0              , the task is to check whether the roots of the equation represented by these constants are reciprocal of each other or not.
Examples: 

Input: A = 2, B = -5, C = 2 
Output: Yes 
Explanation: 
The given quadratic equation is 2x^{2} - 2 = 0
Its roots are (1, 1/1) which are reciprocal of each other.
Input: A = 1, B = -5, C = 6 
Output: No 
Explanation: 
The given quadratic equation is x^{2} - 5x + 6 = 0
Its roots are (2, 3) which are not reciprocal of each other. 

Approach: The idea is to use the concept of quadratic roots to solve the problem. We can formulate the condition required to check whether one root is the reciprocal of the other or not by:  

  1. Let the roots of the equation be \alpha              and \frac{1}{\alpha}              .
  2. The product of the roots of the above equation is given by \alpha              \frac{1}{\alpha}              .
  3. It is known that the product of the roots is C/A. Therefore, the required condition is C = A.

Below is the implementation of the above approach:  

C++

// C++ program to check if roots
// of a Quadratic Equation are
// reciprocal of each other or not
 
#include <iostream>
using namespace std;
 
// Function to check if the roots
// of a quadratic equation are
// reciprocal of each other or not
void checkSolution(int a, int b, int c)
{
    if (a == c)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    int a = 2, b = 0, c = 2;
 
    checkSolution(a, b, c);
 
    return 0;
}

                    

Java

// Java program to check if roots
// of a quadratic equation are
// reciprocal of each other or not
class GFG{
 
// Function to check if the roots 
// of a quadratic equation are
// reciprocal of each other or not
static void checkSolution(int a, int b, int c)
{
    if (a == c)
        System.out.print("Yes");
    else
        System.out.print("No");
}
 
// Driver code
public static void main(String[] args)
{
    int a = 2, b = 0, c = 2;
 
    checkSolution(a, b, c);
}
}
 
// This code is contributed by shubham

                    

Python3

# Python3 program to check if roots
# of a Quadratic Equation are
# reciprocal of each other or not
 
# Function to check if the roots
# of a quadratic equation are
# reciprocal of each other or not
def checkSolution(a, b, c):
 
    if (a == c):
        print("Yes");
    else:
        print("No");
 
# Driver code
a = 2; b = 0; c = 2;
checkSolution(a, b, c);
 
# This code is contributed by Code_Mech

                    

C#

// C# program to check if roots
// of a quadratic equation are
// reciprocal of each other or not
using System;
class GFG{
 
// Function to check if the roots
// of a quadratic equation are
// reciprocal of each other or not
static void checkSolution(int a, int b, int c)
{
    if (a == c)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
 
// Driver code
public static void Main()
{
    int a = 2, b = 0, c = 2;
 
    checkSolution(a, b, c);
}
}
 
// This code is contributed by shivanisinghss2110

                    

Javascript

<script>
 
    // Javascript program to check if roots
    // of a Quadratic Equation are
    // reciprocal of each other or not
     
    // Function to check if the roots
    // of a quadratic equation are
    // reciprocal of each other or not
    function checkSolution(a, b, c)
    {
        if (a == c)
            document.write("Yes");
        else
            document.write("No");
    }
 
    let a = 2, b = 0, c = 2;
  
    checkSolution(a, b, c);
     
</script>

                    

Output
Yes







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

Using Quadratic Formula in python:

We can find the roots of a quadratic equation using the quadratic formula:
x = (-b ± sqrt(b^2 – 4ac)) / 2a
If the roots are reciprocal of each other, then one of the roots is the reciprocal of the other:
x1 * x2 = 1
x2 = 1 / x1
Substituting x2 in terms of x1 in the above quadratic formula:
x1 = (-b ± sqrt(b^2 – 4ac)) / 2a
1 / x1 = (-b ? sqrt(b^2 – 4ac)) / 2a

So, we can find the roots using the quadratic formula and check if they satisfy the above condition to be reciprocal of each other.

  • Calculate the discriminant of the quadratic equation using the formula discriminant = B**2 – 4*A*C.
  • If the discriminant is negative, return False as the quadratic equation has no real roots.
  • Calculate the roots of the quadratic equation using the formula root1 = (-B + math.sqrt(discriminant)) / (2*A) and root2 = (-B – math.sqrt(discriminant)) / (2*A).
  • Check if the product of the roots root1 * root2 is equal to 1.
  • If the product is equal to 1, return True as the roots are reciprocal of each other. Otherwise, return False.

C++

#include <iostream>
#include <cmath>
 
// Function to check if roots are reciprocal of each other
bool areRootsReciprocal(double A, double B, double C) {
    // Calculate the discriminant
    double discriminant = B * B - 4 * A * C;
 
    // If discriminant is negative, roots are imaginary
    if (discriminant < 0) {
        return false;
    }
 
    // Calculate the roots
    double root1 = (-B + sqrt(discriminant)) / (2 * A);
    double root2 = (-B - sqrt(discriminant)) / (2 * A);
 
    // Check if the product of roots is equal to 1
    return root1 * root2 == 1;
}
 
int main() {
    // Example inputs
    double A1 = 2, B1 = -5, C1 = 2;
    double A2 = 1, B2 = -5, C2 = 6;
 
    // Check if roots are reciprocal for the first set of coefficients
    if (areRootsReciprocal(A1, B1, C1)) {
        std::cout << "Yes" << std::endl;
    } else {
        std::cout << "No" << std::endl;
    }
 
    // Check if roots are reciprocal for the second set of coefficients
    if (areRootsReciprocal(A2, B2, C2)) {
        std::cout << "Yes" << std::endl;
    } else {
        std::cout << "No" << std::endl;
    }
 
    return 0;
}

                    

Java

public class Main {
 
    // Function to check if roots are reciprocal of each other
    public static boolean areRootsReciprocal(double A, double B, double C) {
        // Calculate the discriminant
        double discriminant = B * B - 4 * A * C;
 
        // If discriminant is negative, roots are imaginary
        if (discriminant < 0) {
            return false;
        }
 
        // Calculate the roots
        double root1 = (-B + Math.sqrt(discriminant)) / (2 * A);
        double root2 = (-B - Math.sqrt(discriminant)) / (2 * A);
 
        // Check if the product of roots is equal to 1
        return root1 * root2 == 1;
    }
 
    public static void main(String[] args) {
        // Example inputs
        double A1 = 2, B1 = -5, C1 = 2;
        double A2 = 1, B2 = -5, C2 = 6;
 
        // Check if roots are reciprocal for the first set of coefficients
        if (areRootsReciprocal(A1, B1, C1)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
 
        // Check if roots are reciprocal for the second set of coefficients
        if (areRootsReciprocal(A2, B2, C2)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

                    

Python3

import math
 
def are_roots_reciprocal(A, B, C):
    discriminant = B**2 - 4*A*C
    if discriminant < 0:
        return False
    root1 = (-B + math.sqrt(discriminant)) / (2*A)
    root2 = (-B - math.sqrt(discriminant)) / (2*A)
    return root1 * root2 == 1
 
# Example inputs
A1, B1, C1 = 2, -5, 2
A2, B2, C2 = 1, -5, 6
 
# Check if roots are reciprocal of each other
if are_roots_reciprocal(A1, B1, C1):
    print("Yes")
else:
    print("No")
 
if are_roots_reciprocal(A2, B2, C2):
    print("Yes")
else:
    print("No")

                    

C#

using System;
 
class MainClass {
    // Function to check if roots are reciprocal of each
    // other
    static bool AreRootsReciprocal(double A, double B,
                                   double C)
    {
        // Calculate the discriminant
        double discriminant = B * B - 4 * A * C;
 
        // If discriminant is negative, roots are imaginary
        if (discriminant < 0) {
            return false;
        }
 
        // Calculate the roots
        double root1
            = (-B + Math.Sqrt(discriminant)) / (2 * A);
        double root2
            = (-B - Math.Sqrt(discriminant)) / (2 * A);
 
        // Check if the product of roots is equal to 1
        return root1 * root2 == 1;
    }
 
    public static void Main(string[] args)
    {
        // Example inputs
        double A1 = 2, B1 = -5, C1 = 2;
        double A2 = 1, B2 = -5, C2 = 6;
 
        // Check if roots are reciprocal for the first set
        // of coefficients
        if (AreRootsReciprocal(A1, B1, C1)) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
 
        // Check if roots are reciprocal for the second set
        // of coefficients
        if (AreRootsReciprocal(A2, B2, C2)) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
}

                    

Javascript

function areRootsReciprocal(A, B, C) {
    let discriminant = B ** 2 - 4 * A * C;
    if (discriminant < 0) {
        return false;
    }
    let root1 = (-B + Math.sqrt(discriminant)) / (2 * A);
    let root2 = (-B - Math.sqrt(discriminant)) / (2 * A);
    return root1 * root2 === 1;
}
 
// Example inputs
let A1 = 2, B1 = -5, C1 = 2;
let A2 = 1, B2 = -5, C2 = 6;
 
// Check if roots are reciprocal of each other
if (areRootsReciprocal(A1, B1, C1)) {
    console.log("Yes");
} else {
    console.log("No");
}
 
if (areRootsReciprocal(A2, B2, C2)) {
    console.log("Yes");
} else {
    console.log("No");
}

                    

Output
Yes
No







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



Last Updated : 13 Dec, 2023
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