# Find two numbers with sum and product both same as N

Given an integer N, the task is to find two numbers a and b such that a * b = N and a + b = N. Print “NO” if no such numbers are possible.

Examples:

Input: N = 69
Output: a = 67.9851
b = 1.01493

Input: N = 1
Output: NO

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: If observed carefully, we are given with sum and product of roots of a quadratic equation.
If N2 – 4*N < 0 then only imaginary roots are possible for the equation, hence “NO” will be the answer. Else a and b will be:

a = ( N + sqrt( N2 – 4*N ) ) / 2
b = ( N – sqrt( N2 – 4*N ) ) / 2

Below is the implementation of the above approach:

 // C++ program to find a and b // such that a*b=N and a+b=N #include using namespace std;    // Function to return the smallest string void findAandB(double N) {     double val = N * N - 4.0 * N;        // Not possible     if (val < 0) {         cout << "NO";         return;     }        // find a and b     double a = (N + sqrt(val)) / 2.0;     double b = (N - sqrt(val)) / 2.0;        cout << "a = " << a << endl;     cout << "b = " << b << endl; }    // Driver Code int main() {     double N = 69.0;     findAandB(N);     return 0; }

 // Java program to find a and b // such that a*b=N and a+b=N    class GFG{ // Function to return the smallest string static void findAandB(double N) {     double val = N * N - 4.0 * N;        // Not possible     if (val < 0) {         System.out.println("NO");         return;     }        // find a and b     double a = (N + Math.sqrt(val)) / 2.0;     double b = (N - Math.sqrt(val)) / 2.0;        System.out.println("a = "+a);     System.out.println("b = "+b); }    // Driver Code public static void main(String[] args) {     double N = 69.0;     findAandB(N); } } // This Code is contributed by mits

 # Python 3 program to find a and b # such that a*b=N and a+b=N from math import sqrt    # Function to return the  # smallest string def findAandB(N):     val = N * N - 4.0 * N        # Not possible     if (val < 0):         print("NO")         return            # find a and b     a = (N + sqrt(val)) / 2.0     b = (N - sqrt(val)) / 2.0        print("a =", '{0:.6}' . format(a))     print("b =", '{0:.6}' . format(b))    # Driver Code if __name__ == '__main__':     N = 69.0     findAandB(N)        # This code is contributed  # by SURENDRA_GANGWAR

 // C# program to find a and b // such that a*b=N and a+b=N    using System; class GFG { // Function to return the smallest string static void findAandB(double N) { double val = N * N - 4.0 * N;    // Not possible if (val < 0) { Console.WriteLine("NO"); return; }    // find a and b double a = (N + Math.Sqrt(val)) / 2.0; double b = (N - Math.Sqrt(val)) / 2.0;    Console.WriteLine("a = "+a); Console.WriteLine("b = "+b); }    // Driver Code static void Main() { double N = 69.0; findAandB(N); } // This code is contributed by ANKITRAI1 }



Output:
a = 67.9851
b = 1.01493

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Practice Tags :