# Least root of given quadratic equation for value greater than equal to K

Given the constants of quadratic equation F(x) = Ax2 + Bx + C as A, B, and C and an integer K, the task is to find the smallest value of root x such that F(x) ≥ K. If no such values exist then print “-1”. It is given that F(x) is a monotonically increasing function.

Examples:

Input: A = 3, B = 4, C = 5, K = 6
Output: 1
Explanation:
For the given values F(x) = 3x2 + 4x + 5 the minimum value of x is 1, F(x) = 12, which is greater than the given value of K.

Input: A = 3, B = 4, C = 5, K = 150
Output: 7
Explanation:
For the given values F(x) = 3x2 + 4x + 5 the minimum value of x is 7, F(x) = 180, which is greater than the given value of K.

Approach: The idea is to use Binary Search to find the minimum value of x. Below are the steps:

• To get the value equal to or greater than K, the value of x must be in the range [1, sqrt(K)] as this is a quadratic equation.
• Now, basically there is a need to search the appropriate element in the range, so for this binary search is implemented.
• Computing F(mid), where mid is the middle value for the range [1, sqrt(K)]. Now the following three cases are possible:
• If F(mid) ≥ K && F(mid) < K: This mean the current mid is the required answer.
• If F(mid) < K: This means the current value of mid is less than the required value of x.  So, move towards the right, i.e., in the second half as F(x) is an increasing function.

• If F(mid) > K: This means the current value of mid is greater than the required value of x. So, move towards the left, i.e., the first half as F(x) is an increasing function.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate value of ` `// quadratic equation for some x ` `int` `func(``int` `A, ``int` `B, ``int` `C, ``int` `x) ` `{ ` `    ``return` `(A * x * x + B * x + C); ` `} ` ` `  `// Function to calculate the minimum ` `// value of x such that F(x) >= K using ` `// binary search ` `int` `findMinx(``int` `A, ``int` `B, ``int` `C, ``int` `K) ` `{ ` ` `  `    ``// Start and end value for ` `    ``// binary search ` `    ``int` `start = 1; ` `    ``int` `end = ``ceil``(``sqrt``(K)); ` ` `  `    ``// Binary Search ` `    ``while` `(start <= end) { ` `        ``int` `mid = start + (end - start) / 2; ` ` `  `        ``// Computing F(mid) and F(mid-1) ` `        ``int` `x = func(A, B, C, mid); ` `        ``int` `Y = func(A, B, C, mid - 1); ` ` `  `        ``// Checking the three cases ` ` `  `        ``// If F(mid) >= K  and ` `        ``// F(mid-1) < K return mid ` `        ``if` `(x >= K && Y < K) { ` `            ``return` `mid; ` `        ``} ` ` `  `        ``// If F(mid) < K go to mid+1 to end ` `        ``else` `if` `(x < K) { ` `            ``start = mid + 1; ` `        ``} ` ` `  `        ``// If F(mid) > K go to start to mid-1 ` `        ``else` `{ ` `            ``end = mid - 1; ` `        ``} ` `    ``} ` ` `  `    ``// If no such value exist ` `    ``return` `-1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given coefficients of Equations ` `    ``int` `A = 3, B = 4, C = 5, K = 6; ` ` `  `    ``// Find minimum value of x ` `    ``cout << findMinx(A, B, C, K); ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to calculate value of ` `// quadratic equation for some x ` `static` `int` `func(``int` `A, ``int` `B, ``int` `C, ``int` `x) ` `{ ` `    ``return` `(A * x * x + B * x + C); ` `} ` ` `  `// Function to calculate the minimum ` `// value of x such that F(x) >= K using ` `// binary search ` `static` `int` `findMinx(``int` `A, ``int` `B, ``int` `C, ``int` `K) ` `{ ` `     `  `    ``// Start and end value for ` `    ``// binary search ` `    ``int` `start = ``1``; ` `    ``int` `end = (``int``)Math.ceil(Math.sqrt(K)); ` ` `  `    ``// Binary Search ` `    ``while` `(start <= end) ` `    ``{ ` `        ``int` `mid = start + (end - start) / ``2``; ` ` `  `        ``// Computing F(mid) and F(mid-1) ` `        ``int` `x = func(A, B, C, mid); ` `        ``int` `Y = func(A, B, C, mid - ``1``); ` ` `  `        ``// Checking the three cases ` `        ``// If F(mid) >= K and ` `        ``// F(mid-1) < K return mid ` `        ``if` `(x >= K && Y < K)  ` `        ``{ ` `            ``return` `mid; ` `        ``} ` ` `  `        ``// If F(mid) < K go to mid+1 to end ` `        ``else` `if` `(x < K)  ` `        ``{ ` `            ``start = mid + ``1``; ` `        ``} ` ` `  `        ``// If F(mid) > K go to start to mid-1 ` `        ``else`  `        ``{ ` `            ``end = mid - ``1``; ` `        ``} ` `    ``} ` `     `  `    ``// If no such value exist ` `    ``return` `-``1``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given coefficients of Equations ` `    ``int` `A = ``3``, B = ``4``, C = ``5``, K = ``6``; ` `     `  `    ``// Find minimum value of x ` `    ``System.out.println(findMinx(A, B, C, K)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

Output:

```1
```

Time Complexity: O(log(sqrt(K))
Auxiliary Space: O(1)

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Improved By : offbeat