Introduction to Java Programming and Data Structures: Brief Version (11th Global Edition)
Introduction to Java Programming and Data Structures: Brief Version (11th Global Edition)
11th Edition
ISBN: 9780134671710
Author: Y. Daniel Liang
Publisher: PEARSON
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Chapter 3, Problem 3.1PE
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Program Plan Intro

Algebra: solve quadratic equation

Program Plan:

  • Import the required packages.
  • Create a class Exercise
    • Define the main method.
    • Define the scanner input and prompt user to enter the value of a,b,c.
    • Get the input.
    • Calculate the discriminant value.
    • Condition to validate the discriminant value.
    • After validation the value gets of the root gets calculated based on the condition.
    • Display the result of roots.
Program Description Answer

The below program is used to solve the quadratic equation:

Explanation of Solution

Program:

//import the required packages

import java.util.Scanner;

//define the class exercise

public class Exercise

{

public static void main(String[] args)

{

//scanner input

Scanner input = new Scanner(System.in);

/*prompt user to enter the value of the a,b and c*/

System.out.print("Enter a, b, c: ");

//get value of a

double a = input.nextDouble();

//get value of b

double b = input.nextDouble();

//get value of c

double c = input.nextDouble();

//equation to calculate the discriminant

double discriminant = b * b - 4 * a * c;

//condition to validate the discriminant value

if (discriminant < 0)

{

//display result

System.out.println("The equation has no real roots");

}

//condition to validate the discriminant value

else if (discriminant == 0)

{

//equation to calculate the root value

double r1 = -b / (2 * a);

//display result

System.out.println("The equation has one root " + r1);

}

//condition to validate the discriminant value

else

{

// (discriminant > 0)

//equation to calculate the root value

double r1 = (-b + Math.pow(discriminant, 0.5)) / (2 * a);

//equation to calculate the root value

double r2 = (-b - Math.pow(discriminant, 0.5)) / (2 * a);

//display result

System.out.println("The equation has two roots " + r1 + " and " + r2);

}

}

}

Sample Output

Enter a, b, c: 1

2.0

1

The equation has one root -1.0

Additional Output 1:

Enter a, b, c: 1.0

3

1

The equation has two roots -0.3819660112501051 and -2.618033988749895

Additional Output 2:

Enter a, b, c: 1

2

3

The equation has no real roots

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Chapter 3 Solutions

Introduction to Java Programming and Data Structures: Brief Version (11th Global Edition)

Chapter 3.5, Problem 3.5.3CPChapter 3.6, Problem 3.6.1CPChapter 3.6, Problem 3.6.2CPChapter 3.6, Problem 3.6.3CPChapter 3.6, Problem 3.6.4CPChapter 3.7, Problem 3.7.1CPChapter 3.7, Problem 3.7.2CPChapter 3.9, Problem 3.9.1CPChapter 3.10, Problem 3.10.1CPChapter 3.10, Problem 3.10.2CPChapter 3.10, Problem 3.10.3CPChapter 3.10, Problem 3.10.4CPChapter 3.10, Problem 3.10.5CPChapter 3.10, Problem 3.10.6CPChapter 3.10, Problem 3.10.7CPChapter 3.10, Problem 3.10.8CPChapter 3.10, Problem 3.10.9CPChapter 3.10, Problem 3.10.10CPChapter 3.10, Problem 3.10.11CPChapter 3.11, Problem 3.11.1CPChapter 3.12, Problem 3.12.1CPChapter 3.13, Problem 3.13.1CPChapter 3.13, Problem 3.13.2CPChapter 3.13, Problem 3.13.3CPChapter 3.13, Problem 3.13.4CPChapter 3.13, Problem 3.13.5CPChapter 3.14, Problem 3.14.1CPChapter 3.14, Problem 3.14.2CPChapter 3.14, Problem 3.14.3CPChapter 3.14, Problem 3.14.4CPChapter 3.15, Problem 3.15.1CPChapter 3.15, Problem 3.15.2CPChapter 3.15, Problem 3.15.3CPChapter 3.15, Problem 3.15.4CPChapter 3, Problem 3.1PEChapter 3, Problem 3.2PEChapter 3, Problem 3.3PEChapter 3, Problem 3.4PEChapter 3, Problem 3.5PEChapter 3, Problem 3.6PEChapter 3, Problem 3.7PEChapter 3, Problem 3.8PEChapter 3, Problem 3.9PEChapter 3, Problem 3.10PEChapter 3, Problem 3.11PEChapter 3, Problem 3.12PEChapter 3, Problem 3.13PEChapter 3, Problem 3.14PEChapter 3, Problem 3.15PEChapter 3, Problem 3.16PEChapter 3, Problem 3.17PEChapter 3, Problem 3.18PEChapter 3, Problem 3.19PEChapter 3, Problem 3.20PEChapter 3, Problem 3.21PEChapter 3, Problem 3.22PEChapter 3, Problem 3.23PEChapter 3, Problem 3.24PEChapter 3, Problem 3.25PEChapter 3, Problem 3.26PEChapter 3, Problem 3.27PEChapter 3, Problem 3.28PEChapter 3, Problem 3.29PEChapter 3, Problem 3.30PEChapter 3, Problem 3.31PEChapter 3, Problem 3.32PEChapter 3, Problem 3.33PEChapter 3, Problem 3.34PE
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