Homework Help is Here – Start Your Trial Now!
Engineering
Computer Engineering
Design a class named QuadraticEquation for solving a quadratic equation which is of the form ax²+bx+c=0.

Design a class named QuadraticEquation for solving a quadratic equation which is of the form ax²+bx+c=0.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
See similar textbooks
icon
Related questions
Question

Please solve it in Java...

 

Design a class named QuadraticEquation for solving a quadratic equation which is of the form ax²+bx+c=0. QuadraticEquation - a: double - b: double - c: double + QuadraticEquation(a: double, b: double, c: double) - getDiscriminant(): double + evaluateDiscriminant(): void + getRoot1(): double + getRoot2(0: double + toString(): String The class should contain: Private data fields a , b , and c that represent three coefficients. • A single constructor for the arguments for a , b, and c. A private method named getDiscriminant() that returns the discriminant, which is the value of b?-4 ac. A public method named evaluateDiscriminant() which prints out the following information using the value obtained from getDiscriminant(): o getDiscriminant() > 0, "Two distinct solutions" getDiscriminant() o getDiscriminant() < 0, “No real solutions" The public methods named getRoot1() and getRoot2() return the two roots of the equation. They return the following, respectively: = 0, "One unique solution" -b + Vb - 4ac -b – VB - 4ac and r2 = 2a 2a However, if no real solutions exist, print out the message "Not possible to calculate solutions since discriminant is negative". You may have to return an arbitrary value in this case. The toString() method returns the details in the following tabular format: Faculty: SAS3 CSE215L 1 Coefficient | Value | value of a upto 3 decimal places a b | value of b upto 3 decimal places | value of c upto 3 decimal places In your driver class, construct three objects of QuadraticEquation. Use Scanner class to get the values of the coefficients for each object. Call the evaluateDiscriminant() method on each of the objects. Then, call the getRoot1() and getRoot2() methods on each of the objects and display the results. ||
Transcribed Image Text:Design a class named QuadraticEquation for solving a quadratic equation which is of the form ax²+bx+c=0. QuadraticEquation - a: double - b: double - c: double + QuadraticEquation(a: double, b: double, c: double) - getDiscriminant(): double + evaluateDiscriminant(): void + getRoot1(): double + getRoot2(0: double + toString(): String The class should contain: Private data fields a , b , and c that represent three coefficients. • A single constructor for the arguments for a , b, and c. A private method named getDiscriminant() that returns the discriminant, which is the value of b?-4 ac. A public method named evaluateDiscriminant() which prints out the following information using the value obtained from getDiscriminant(): o getDiscriminant() > 0, "Two distinct solutions" getDiscriminant() o getDiscriminant() < 0, “No real solutions" The public methods named getRoot1() and getRoot2() return the two roots of the equation. They return the following, respectively: = 0, "One unique solution" -b + Vb - 4ac -b – VB - 4ac and r2 = 2a 2a However, if no real solutions exist, print out the message "Not possible to calculate solutions since discriminant is negative". You may have to return an arbitrary value in this case. The toString() method returns the details in the following tabular format: Faculty: SAS3 CSE215L 1 Coefficient | Value | value of a upto 3 decimal places a b | value of b upto 3 decimal places | value of c upto 3 decimal places In your driver class, construct three objects of QuadraticEquation. Use Scanner class to get the values of the coefficients for each object. Call the evaluateDiscriminant() method on each of the objects. Then, call the getRoot1() and getRoot2() methods on each of the objects and display the results. ||
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Computer Networking: A Top-Down Approach (7th Edi…
Computer Networking: A Top-Down Approach (7th Edi…
Computer Engineering
ISBN:
9780133594140
Author:
James Kurose, Keith Ross
Publisher:
PEARSON
Computer Organization and Design MIPS Edition, Fi…
Computer Organization and Design MIPS Edition, Fi…
Computer Engineering
ISBN:
9780124077263
Author:
David A. Patterson, John L. Hennessy
Publisher:
Elsevier Science
Network+ Guide to Networks (MindTap Course List)
Network+ Guide to Networks (MindTap Course List)
Computer Engineering
ISBN:
9781337569330
Author:
Jill West, Tamara Dean, Jean Andrews
Publisher:
Cengage Learning
Concepts of Database Management
Concepts of Database Management
Computer Engineering
ISBN:
9781337093422
Author:
Joy L. Starks, Philip J. Pratt, Mary Z. Last
Publisher:
Cengage Learning
Prelude to Programming
Prelude to Programming
Computer Engineering
ISBN:
9780133750423
Author:
VENIT, Stewart
Publisher:
Pearson Education
Sc Business Data Communications and Networking, T…
Sc Business Data Communications and Networking, T…
Computer Engineering
ISBN:
9781119368830
Author:
FITZGERALD
Publisher:
WILEY
SEE MORE TEXTBOOKS