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Find minimum positive integer x such that a(x^2) + b(x) + c >= k
• Difficulty Level : Basic
• Last Updated : 23 Apr, 2019

Given four integers a, b, c and k. The task is to find the minimum positive value of x such that ax2 + bx + c ≥ k.

Examples:

Input: a = 3, b = 4, c = 5, k = 6
Output: 1
For x = 0, a * 0 + b * 0 + c = 5 < 6
For x = 1, a * 1 + b * 1 + c = 3 + 4 + 5 = 12 > 6

Input: a = 2, b = 7, c = 6, k = 3
Output: 0

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

Approach: The idea is to use binary search. The lower limit for our search will be 0 since x has to be minimum positive integer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the minimum positive``// integer satisfying the given equation``int` `MinimumX(``int` `a, ``int` `b, ``int` `c, ``int` `k)``{``    ``int` `x = INT_MAX;`` ` `    ``if` `(k <= c)``        ``return` `0;`` ` `    ``int` `h = k - c;``    ``int` `l = 0;`` ` `    ``// Binary search to find the value of x``    ``while` `(l <= h) {``        ``int` `m = (l + h) / 2;``        ``if` `((a * m * m) + (b * m) > (k - c)) {``            ``x = min(x, m);``            ``h = m - 1;``        ``}``        ``else` `if` `((a * m * m) + (b * m) < (k - c))``            ``l = m + 1;``        ``else``            ``return` `m;``    ``}`` ` `    ``// Return the answer``    ``return` `x;``}`` ` `// Driver code``int` `main()``{``    ``int` `a = 3, b = 2, c = 4, k = 15;``    ``cout << MinimumX(a, b, c, k);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``     ` `// Function to return the minimum positive``// integer satisfying the given equation``static` `int` `MinimumX(``int` `a, ``int` `b, ``int` `c, ``int` `k)``{``    ``int` `x = Integer.MAX_VALUE;`` ` `    ``if` `(k <= c)``        ``return` `0``;`` ` `    ``int` `h = k - c;``    ``int` `l = ``0``;`` ` `    ``// Binary search to find the value of x``    ``while` `(l <= h) ``    ``{``        ``int` `m = (l + h) / ``2``;``        ``if` `((a * m * m) + (b * m) > (k - c)) ``        ``{``            ``x = Math.min(x, m);``            ``h = m - ``1``;``        ``}``        ``else` `if` `((a * m * m) + (b * m) < (k - c))``            ``l = m + ``1``;``        ``else``            ``return` `m;``    ``}`` ` `    ``// Return the answer``    ``return` `x;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``3``, b = ``2``, c = ``4``, k = ``15``;``    ``System.out.println(MinimumX(a, b, c, k));``}``}`` ` `// This code is contributed by Code_Mech. `

## Python3

 `# Python3 implementation of the approach`` ` `# Function to return the minimum positive``# integer satisfying the given equation``def` `MinimumX(a, b, c, k):`` ` `    ``x ``=` `10``*``*``9`` ` `    ``if` `(k <``=` `c):``        ``return` `0`` ` `    ``h ``=` `k ``-` `c``    ``l ``=` `0`` ` `    ``# Binary search to find the value of x``    ``while` `(l <``=` `h):``        ``m ``=` `(l ``+` `h) ``/``/` `2``        ``if` `((a ``*` `m ``*` `m) ``+` `(b ``*` `m) > (k ``-` `c)):``            ``x ``=` `min``(x, m)``            ``h ``=` `m ``-` `1`` ` `        ``elif` `((a ``*` `m ``*` `m) ``+` `(b ``*` `m) < (k ``-` `c)):``            ``l ``=` `m ``+` `1``        ``else``:``            ``return` `m`` ` `    ``# Return the answer``    ``return` `x`` ` `# Driver code``a, b, c, k ``=` `3``, ``2``, ``4``, ``15``print``(MinimumX(a, b, c, k))`` ` `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach``using` `System;`` ` `class` `GFG``{``     ` `// Function to return the minimum positive``// integer satisfying the given equation``static` `int` `MinimumX(``int` `a, ``int` `b, ``int` `c, ``int` `k)``{``    ``int` `x = ``int``.MaxValue;`` ` `    ``if` `(k <= c)``        ``return` `0;`` ` `    ``int` `h = k - c;``    ``int` `l = 0;`` ` `    ``// Binary search to find the value of x``    ``while` `(l <= h) ``    ``{``        ``int` `m = (l + h) / 2;``        ``if` `((a * m * m) + (b * m) > (k - c)) ``        ``{``            ``x = Math.Min(x, m);``            ``h = m - 1;``        ``}``        ``else` `if` `((a * m * m) + (b * m) < (k - c))``            ``l = m + 1;``        ``else``            ``return` `m;``    ``}`` ` `    ``// Return the answer``    ``return` `x;``}`` ` `// Driver code``public` `static` `void` `Main()``{``    ``int` `a = 3, b = 2, c = 4, k = 15;``    ``Console.Write(MinimumX(a, b, c, k));``}``}`` ` `// This code is contributed by Akanksha Rai`

## PHP

 ` (``\$k` `- ``\$c``))``        ``{ ``            ``\$x` `= min(``\$x``, ``\$m``); ``            ``\$h` `= ``\$m` `- 1; ``        ``} ``        ``else` `if` `((``\$a` `* ``\$m` `* ``\$m``) + ``                 ``(``\$b` `* ``\$m``) < (``\$k` `- ``\$c``)) ``            ``\$l` `= ``\$m` `+ 1; ``        ``else``            ``return` `\$m``; ``    ``} `` ` `    ``// Return the answer ``    ``return` `\$x``; ``} `` ` `// Driver code ``\$a` `= 3; ``\$b` `= 2; ``\$c` `= 4; ``\$k` `= 15; `` ` `echo` `MinimumX(``\$a``, ``\$b``, ``\$c``, ``\$k``);`` ` `// This code is contributed by Ryuga``?>`
Output:
```2
```

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