# 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
```

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