Program to find number of solutions in Quadratic Equation (original) (raw)

Last Updated : 30 Aug, 2022

Given an equation ax^{2}\pm bx\pm c=0 with value a, b, and c, where a and b is any value and c is constant, find how many solutions thus this quadratic equation have?
Examples:

Input : 2x^{2}+5x+2=0Output : 2 solutionsInput : x^{2}+x+1=0Output : no solution

Solution:
To check whether the equation has a solution or not, quadratic formula for discriminant is used.

The formula is given as, b^{2}-4ac

Respective conditions are given as,

if the discriminant is positive b^{2}-4ac> 0 , then the quadratic equation has two solutions.
if the discriminant is equal b^{2}-4ac= 0 , then the quadratic equation has one solution.
if the discriminant is negative b^{2}-4ac< 0 , then the quadratic equation has no solution.

Programs:

C++ `

// C++ Program to find the solutions of specified equations #include using namespace std;

// Method to check for solutions of equations void checkSolution(int a, int b, int c) {

// If the expression is greater than 0, then 2 solutions
if (((b * b) - (4 * a * c)) > 0)
    cout << "2 solutions";

// If the expression is equal 0, then 2 solutions
else if (((b * b) - (4 * a * c)) == 0)
    cout << "1 solution";

// Else no solutions
else
    cout << "No solutions";

}

int main() { int a = 2, b = 5, c = 2; checkSolution(a, b, c); return 0; }

Java

// Java Program to find the solutions of specified equations public class GFG {

// Method to check for solutions of equations
static void checkSolution(int a, int b, int c)
{

    // If the expression is greater than 0, 
    // then 2 solutions
    if (((b * b) - (4 * a * c)) > 0)
        System.out.println("2 solutions");

    // If the expression is equal 0, then 2 solutions
    else if (((b * b) - (4 * a * c)) == 0)
        System.out.println("1 solution");

    // Else no solutions
    else
        System.out.println("No solutions");
}

// Driver Code
public static void main(String[] args)
{
    int a = 2, b = 5, c = 2;
    checkSolution(a, b, c);
}

}

Python3

Python3 Program to find the

solutions of specified equations

function to check for

solutions of equations

def checkSolution(a, b, c) :

# If the expression is greater
# than 0, then 2 solutions 
if ((b * b) - (4 * a * c)) > 0 :
    print("2 solutions")

# If the expression is equal 0, 
# then 1 solutions
elif ((b * b) - (4 * a * c)) == 0 :
    print("1 solution")

# Else no solutions 
else :
    print("No solutions")

Driver code

if name == "main" :

a, b, c = 2, 5, 2
checkSolution(a, b, c)

This code is contributed

by ANKITRAI1

C#

// C# Program to find the solutions // of specified equations using System; class GFG {

// Method to check for solutions of equations static void checkSolution(int a, int b, int c) {

// If the expression is greater 
// than 0, then 2 solutions
if (((b * b) - (4 * a * c)) > 0)
    Console.WriteLine("2 solutions");

// If the expression is equal to 0,
// then 2 solutions
else if (((b * b) - (4 * a * c)) == 0)
    Console.WriteLine("1 solution");

// Else no solutions
else
    Console.WriteLine("No solutions");

}

// Driver Code public static void Main() { int a = 2, b = 5, c = 2; checkSolution(a, b, c); } }

// This code is contributed by inder_verma

PHP

b,b, b,c) { // If the expression is greater // than 0, then 2 solutions if ((($b * b)−(4∗b) - (4 * b)−(4∗a * $c)) > 0) echo "2 solutions"; // If the expression is equal 0, // then 2 solutions else if ((($b * $b) - (4 * a∗a * a∗c)) == 0) echo "1 solution"; // Else no solutions else echo"No solutions"; } // Driver Code a=2;a = 2; a=2;b = 5; $c = 2; checkSolution($a, b,b, b,c); // This code is contributed // by inder_verma ?>

JavaScript

Output:

2 solutions

Time Complexity: O(1)

Auxiliary Space: O(1)