Write a program to find the square root of the quadratic equation?

Answer 1

import java.io.*;

public class quadratic

{

private String str;

public quadratic()

{

str = "";

}

public quadratic(String str)

{

this.str = str;

}

public void accept()throws IOException

{

BufferedReader br= new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter the equation in terms of x \nEg: 1x^2+4x+5=0 \nAnd you will receive the possible results of x");

str= br.readLine();

}

public void solve()

{

String A[]=new String [8];

int l=str.length();

int j=0;

char c=' ';

String w="";

for(int i=0;i<7;i++)

{

while(true)

{

c=str.charAt(j);

if(c!='+'&&c!='-'&&c!='=')

w=w+c;

else

{

if(i==4)

A[i]="0"+w;

else

A[i]=w;

i++;

if(i>6)

break;

A[i]=" "+c;

j++;

break;

}

j++;

if(j>=l)

break;

}

w="";

}

System.out.println(A[0]);

System.out.println(A[2]);

System.out.println(A[6]);

int len=A[0].length();

String M=A[0].substring(0,(len-3));

len=A[2].length();

String B=A[2].substring(0,len-1);

int a=Integer.parseInt(M);

int b=Integer.parseInt(B);

int C=Integer.parseInt(A[4]);

if(A[3].equals(" -"))

C=-C;

if(A[1].equals(" -"))

b=-b;

double D=(double)Math.sqrt(b*b-4*a*C);

double ans1=(-b+D)/(2*a);

double ans2=(-b-D)/(2*a);

System.out.println("The values of x are: "+ans1+" and "+ans2);

}

public static void main(String args[])throws IOException

{

quadratic ob = new quadratic();

ob.accept();

ob.solve();

}

}

Answer 2

import java.util.*;

public class quadratic

{

public static void main(String[] args)

{

Scanner input = new Scanner(System.in);

String equation = "( ax^2 + bx + c = 0 1 )";

System.out.println("The equation is " + equation);

System.out.println("Please enter the values of coefficients a, b, c, separated by spaces");

System.out.println("a = "); double a = input.nextDouble();

System.out.println("b = "); double b = input.nextDouble();

System.out.println("c = "); double c = input.nextDouble();

System.out.println();

double discriminant = Math.pow(b,2) - 4*a*c;

double root1 = (-b + Math.sqrt(discriminant))/(2*a);

double root2 = (-b - Math.sqrt(discriminant))/(2*a);

System.out.println(" " + equation);

System.out.println("The coefficients you entered were " + "\nA = " + a + "\nB = " + b + "\nC = " + c);

System.out.println();

if(discriminant > 0)

{

System.out.println("Root1: " + root1);

System.out.println("Root2: " + root2);

}

else if(discriminant == 0)

{

System.out.println("Root: " + root1);

}

else

{

double root1complex = -b/(2*a);

double root1complex2 = Math.sqrt(-discriminant)/(2*a);

System.out.println(root1complex + " + " + root1complex2 + " i ");

System.out.println(" " + " and ");

System.out.println(root1complex + " - " + root1complex2 + " i ");

}

}

}

Answer 3

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<process.h>

void main()

{

float a,b,c,x1,x2,disc;

clrscr();

printf("Enter the co-efficients\n");

scanf("%f%f%f",&a,&b,&c);

disc=b*b-4*a*c;/*to find discriminant*/

if(disc>0)/*distinct roots*/

{

x1=(-b+sqrt(disc))/(2*a);

x2=(-b-sqrt(disc))/(2*a);

printf("The roots are distinct\n");

exit(0);

}

if(disc==0)/*Equal roots*/

{

x1=x2=-b/(2*a);

printf("The roots are equal\n");

printf("x1=%f\nx2=%f\n",x1,x2);

exit(0);

}

x1=-b/(2*a);/*complex roots*/

x2=sqrt(fabs(disc))/(2*a);

printf("The roots are complex\n");

printf("The first root=%f+i%f\n",x1,x2);

printf("The second root=%f-i%f\n",x1,x2);

getch();

}

Answer 4

class Quadratic

{

public static void findRoots(double a, double b, double c)

{

double disc;

disc = (b * b) - (4 * a * c);

if ( disc > 0 )

{

double Root1 = (- b + Math.sqrt(disc)) / (2 * a);

double Root2 = (- b - Math.sqrt(disc)) / (2 * a);

System.out.println("The roots are unequal and irrational " + Root1 + " and " + Root2);

}

else if ( disc == 0 )

{

double root = - b / (2 * a);

System.out.println("The roots are equal and rational " + root + " and " + root );

}

else if ( disc < 0 )

{

System.out.println("The roots are imaginary");

}

} // function is over

} // class is over