Given the roots of a quadratic equation A and B, the task is to find the equation.
Note: The given roots are integral.
Examples:
Input: A = 2, B = 3
Output: x^2 – (5x) + (6) = 0
x2 – 5x + 6 = 0
x2 -3x -2x + 6 = 0
x(x – 3) – 2(x – 3) = 0
(x – 3) (x – 2) = 0
x = 2, 3
Input: A = 5, B = 10
Output: x^2 – (15x) + (50) = 0
Approach: If the roots of a quadratic equation ax2 + bx + c = 0 are A and B then it known that
A + B = – b / a and A * B = c * a.
Now, ax2 + bx + c = 0 can be written as
x2 + (b / a)x + (c / a) = 0 (Since, a != 0)
x2 – (A + B)x + (A * B) = 0, [Since, A + B = -b * a and A * B = c * a]
i.e. x2 – (Sum of the roots)x + Product of the roots = 0
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findEquation( int a, int b)
{
int sum = (a + b);
int product = (a * b);
cout << "x^2 - (" << sum << "x) + ("
<< product << ") = 0" ;
}
int main()
{
int a = 2, b = 3;
findEquation(a, b);
return 0;
}
|
Java
class GFG
{
static void findEquation( int a, int b)
{
int sum = (a + b);
int product = (a * b);
System.out.println( "x^2 - (" + sum +
"x) + (" + product + ") = 0" );
}
public static void main(String args[])
{
int a = 2 , b = 3 ;
findEquation(a, b);
}
}
|
Python3
def findEquation(a, b):
summ = (a + b)
product = (a * b)
print ( "x^2 - (" , summ,
"x) + (" , product, ") = 0" )
a = 2
b = 3
findEquation(a, b)
|
C#
using System;
class GFG
{
static void findEquation( int a, int b)
{
int sum = (a + b);
int product = (a * b);
Console.WriteLine( "x^2 - (" + sum +
"x) + (" + product + ") = 0" );
}
public static void Main()
{
int a = 2, b = 3;
findEquation(a, b);
}
}
|
Javascript
<script>
function findEquation(a, b)
{
var sum = (a + b);
var product = (a * b);
document.write( "x^2 - (" + sum +
"x) + (" + product +
") = 0" );
}
var a = 2, b = 3;
findEquation(a, b);
</script>
|
Output: x^2 - (5x) + (6) = 0
Time Complexity: O(1)
Auxiliary Space: O(1)