C Program to Find Solution of Quadratic Equation
Given a quadratic equation in the form ax2 + bx + c, find roots of it.
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Examples :
Input : a = 1, b = -2, c = 1 Output : Roots are real and same 1 Input : a = 1, b = 7, c = 12 Output : Roots are real and different -3, -4 Input : a = 1, b = 1, c = 1 Output : Roots are complex -0.5 + i1.73205 -0.5 - i1.73205
Below is the direct formula for finding roots of the quadratic equation.
There are the following important cases.
If b*b < 4*a*c, then roots are complex (not real). For example roots of x2 + x + 1, roots are -0.5 + i1.73205 and -0.5 - i1.73205 If b*b == 4*a*c, then roots are real and both roots are same. For example, roots of x2 - 2x + 1 are 1 and 1 If b*b > 4*a*c, then roots are real and different. For example, roots of x2 - 7x - 12 are 3 and 4
Below is the implementation of the above formula.
C
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
void findRoots( int a, int b, int c)
{
if (a == 0) {
printf ( "Invalid" );
return ;
}
int d = b * b - 4 * a * c;
double sqrt_val = sqrt ( abs (d));
if (d > 0) {
printf ( "Roots are real and different \n" );
printf ( "%f\n%f" , ( double )(-b + sqrt_val) / (2 * a),
( double )(-b - sqrt_val) / (2 * a));
}
else if (d == 0) {
printf ( "Roots are real and same \n" );
printf ( "%f" , -( double )b / (2 * a));
}
else
{
printf ( "Roots are complex \n" );
printf ( "%f + i%f\n%f - i%f" , -( double )b / (2 * a),
sqrt_val/(2 * a), -( double )b / (2 * a), sqrt_val/(2 * a);
}
}
int main()
{
int a = 1, b = -7, c = 12;
findRoots(a, b, c);
return 0;
}
C++
#include <bits/stdc++.h>
using namespace std;
void findRoots( int a, int b, int c)
{
if (a == 0) {
cout << "Invalid" ;
return ;
}
int d = b * b - 4 * a * c;
double sqrt_val = sqrt ( abs (d));
if (d > 0) {
cout << "Roots are real and different \n" ;
cout << ( double )(-b + sqrt_val) / (2 * a) << "\n"
<< ( double )(-b - sqrt_val) / (2 * a);
}
else if (d == 0) {
cout << "Roots are real and same \n" ;
cout << -( double )b / (2 * a);
}
else
{
cout << "Roots are complex \n" ;
cout << -( double )b / (2 * a) << " + i" << sqrt_val
<< "\n"
<< -( double )b / (2 * a) << " - i" << sqrt_val;
}
}
int main()
{
int a = 1, b = -7, c = 12;
findRoots(a, b, c);
return 0;
}
Java
import java.io.*;
import static java.lang.Math.*;
class Quadratic {
static void findRoots( int a, int b, int c)
{
if (a == 0 ) {
System.out.println( "Invalid" );
return ;
}
int d = b * b - 4 * a * c;
double sqrt_val = sqrt(abs(d));
if (d > 0 ) {
System.out.println(
"Roots are real and different \n" );
System.out.println(
( double )(-b + sqrt_val) / ( 2 * a) + "\n"
+ ( double )(-b - sqrt_val) / ( 2 * a));
}
else if (d == 0 ) {
System.out.println(
"Roots are real and same \n" );
System.out.println(-( double )b / ( 2 * a) + "\n"
+ -( double )b / ( 2 * a));
}
else
{
System.out.println( "Roots are complex \n" );
System.out.println(-( double )b / ( 2 * a) + " + i"
+ sqrt_val + "\n"
+ -( double )b / ( 2 * a)
+ " - i" + sqrt_val);
}
}
public static void main(String args[])
{
int a = 1 , b = - 7 , c = 12 ;
findRoots(a, b, c);
}
}
Python3
import math
def findRoots(a, b, c):
if a = = 0 :
print ( "Invalid" )
return - 1
d = b * b - 4 * a * c
sqrt_val = math.sqrt( abs (d))
if d > 0 :
print ( "Roots are real and different " )
print (( - b + sqrt_val) / ( 2 * a))
print (( - b - sqrt_val) / ( 2 * a))
elif d = = 0 :
print ( "Roots are real and same" )
print ( - b / ( 2 * a))
else :
print ( "Roots are complex" )
print ( - b / ( 2 * a), " + i" , sqrt_val)
print ( - b / ( 2 * a), " - i" , sqrt_val)
a = 1
b = - 7
c = 12
findRoots(a, b, c)
C#
using System;
class Quadratic {
void findRoots( int a, int b, int c)
{
if (a == 0) {
Console.Write( "Invalid" );
return ;
}
int d = b * b - 4 * a * c;
double sqrt_val = Math.Abs(d);
if (d > 0) {
Console.Write(
"Roots are real and different \n" );
Console.Write(
( double )(-b + sqrt_val) / (2 * a) + "\n"
+ ( double )(-b - sqrt_val) / (2 * a));
}
else {
Console.Write( "Roots are complex \n" );
Console.Write(-( double )b / (2 * a) + " + i"
+ sqrt_val + "\n"
+ -( double )b / (2 * a) + " - i"
+ sqrt_val);
}
}
public static void Main()
{
Quadratic obj = new Quadratic();
int a = 1, b = -7, c = 12;
obj.findRoots(a, b, c);
}
}
PHP
<?php
function findRoots( $a , $b , $c )
{
if ( $a == 0)
{
echo "Invalid" ;
return ;
}
$d = $b * $b - 4 * $a * $c ;
$sqrt_val = sqrt( abs ( $d ));
if ( $d > 0)
{
echo "Roots are real and " .
"different \n" ;
echo (- $b + $sqrt_val ) / (2 * $a ) , "\n" ,
(- $b - $sqrt_val ) / (2 * $a );
}
else if ( $d == 0)
{
echo "Roots are real and same \n" ;
echo - $b / (2 * $a );
}
else
{
echo "Roots are complex \n" ;
echo - $b / (2 * $a ) , " + i" ,
$sqrt_val , "\n" , - $b / (2 * $a ),
" - i" , $sqrt_val ;
}
}
$a = 1; $b = -7 ; $c = 12;
findRoots( $a , $b , $c );
?>
Javascript
<script>
function findRoots(a, b, c)
{
if (a == 0) {
document.write( "Invalid" );
return ;
}
let d = b * b - 4 * a * c;
let sqrt_val = Math.sqrt(Math.abs(d));
if (d > 0) {
document.write(
"Roots are real and different \n" + "<br/>" );
document.write(
(-b + sqrt_val) / (2 * a) + "<br/>"
+ (-b - sqrt_val) / (2 * a));
}
else if (d == 0) {
document.write(
"Roots are real and same \n" + "<br/>" );
document.write(-b / (2 * a) + "<br/>"
+ -b / (2 * a)) ;
}
else
{
document.write( "Roots are complex \n" );
document.write(-b / (2 * a) + " + i"
+ sqrt_val + "<br/>"
+ -b / (2 * a)
+ " - i" + sqrt_val);
}
}
let a = 1, b = -7, c = 12;
findRoots(a, b, c);
</script>
Output
Roots are real and different 4.000000 3.000000
This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or hare more information about the topic discussed above.
C Program to Find Solution of Quadratic Equation
Source: https://www.geeksforgeeks.org/program-to-find-the-roots-of-quadratic-equation/