Quadratic equation whose roots are reciprocal to the roots of given equation

Given three integers A, B, and C representing the coefficients of a quadratic equation Ax² + Bx + C = 0, the task is to find the quadratic equation whose roots are reciprocal to the roots of the given equation.

Examples:

Input: A = 1, B = -5, C = 6 
Output: (6)x^2 +(-5)x + (1) = 0
Explanation: 
The given quadratic equation x² – 5x + 6 = 0.
Roots of the above equation are 2, 3.
Reciprocal of these roots are 1/2, 1/3.
Therefore, the quadratic equation with these reciprocal roots is 6x² – 5x + 1 = 0.

Input: A = 1, B = -7, C = 12
Output: (12)x^2 +(-7)x + (1) = 0

Approach: The idea is to use the concept of quadratic roots to solve the problem. Follow the steps below to solve the problem:

• Consider the roots of the equation Ax² + Bx + C = 0 to be p, q.
• The product of the roots of the above equation is given by p * q = C / A.
• The sum of the roots of the above equation is given by p + q = -B / A.
• Therefore, the reciprocals of the roots are 1/p, 1/q.
• The product of these reciprocal roots is 1/p * 1/q = A / C.
• The sum of these reciprocal roots is 1/p + 1/q = -B / C.
• If the sum and product of roots is known, the quadratic equation can be x² – (Sum of the roots)x + (Product of the roots) = 0.
• On solving the above equation, quadratic equation becomes Cx² + Bx + A = 0.

Below is the implementation of the above approach: 

(6)x^2 +(-5)x + (1) = 0

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