Question

Find Roots of a Quadratic Equation

• Step 1: Analyze the Problem
  • Accept three coefficients a, b, and c (all of data type double) of a quadratic equation: ax²+bx+c=0
  • Output real roots (double) of the equation

• Step 2: Develop a Solution
  • Check for degenerate case (user enters 0 for both a and b), no solution
  • Check if the equation is linear (user enters 0 for a), x = -c/b
  • Calculate the discriminant = b²-4ac (obviously, the data type will be double)
  • Check if discriminant > 0, two real roots
  • Check if discriminant < 0, two complex roots
  • Check if discriminant = 0, repeated roots, x = -b/(2a)

Hand calculate and write down the results for the following different cases:

• Real roots case: a = 1, b = 2, c = -35 -> x = 5, -7
• Degenerate roots case: a = b = 0, c = 16
• Repeated roots case: a = 1, b = -14, c = 49 -> x = ?, ?
• Imaginary roots case: a = 2, b = -14, c = 25
• Linear equation case: a = 0, b = -14, c = 28 -> x = ?
• Imaginary roots case: a = 2, b = 0, c = 25 -> x = ?, ?

• Step 3: Code the Solution – Program included in the next page.

• Step 4: Test and Correct the Program for at least the following 5 cases
  • Real roots case: a = 1, b = 2, c = -35
  • Degenerate roots case: a = b = 0, c = 16
  • Repeated roots case: a = 1, b = -14, c = 49
  • Imaginary roots case: a = 2, b = -14, c = 25
  • Linear equation case: a = 0, b = -14, c = 28
  • Imaginary roots case: a = 2, b = 0, c = 25

(a) Compile and run the following program using test conditions outlined above. [5+2*5 points]

#include <iostream>

#include <cmath>

using namespace std;

// this program solves for the roots of a quadratic equation

int main()

{

double a, b, c, disc, root1, root2;

cout << "This program calculates the roots of a\n";

cout << "\tquadratic equation of the form\n";

cout << "\t\t 2\n";

cout << "\t\tax + bx + c = 0\n\n";

cout << "Please enter values for a, b, and c: ";

cin >> a >> b >> c;

if ( a == 0.0 && b == 0.0)

cout << "The equation is degenerate and has no roots.\n";

else if (a == 0.0)

cout << "The equation has the single root x = " << -c/b << endl;

else {

disc = pow(b,2.0) - 4 * a * c;    // calculate discriminant

if (disc > 0.0) {

disc = sqrt(disc);

root1 = (-b + disc) / (2 * a);

root2 = (-b - disc) / (2 * a);

cout << "The two real roots are " << root1 << " and " << root2

<< endl;

}

else if (disc < 0.0)

cout << "Both roots are imaginary.\n";

else

cout << "Both roots are equal to " << -b / (2 * a) << endl;

}

return 0;

}

Copy and paste your Console Debug window outputs here

(b) Modify the program to output complex roots in a format like “4.0 + 3.0i”.        [5 + 5 points]

HINT: For disc < 0.0, you may separately compute real and imaginary parts prior to displaying them.

Copy and paste your modified code here:

Copy and paste Console Debug window output for the complex conjugate roots here:

Answer 1

Solution :

Console output for all the 6 cases :

a)

1.

This program calculates the roots of a
        quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 1 2 -35
The two real roots are 5 and -7

2.

This program calculates the roots of a
        quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 0 0 16
The equation is degenerate and has no roots.

3.

This program calculates the roots of a
quadratic equation of the form
2
ax + bx + c = 0

Please enter values for a, b, and c: 1 -14 49
Both roots are equal to 7

4.

This program calculates the roots of a
quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 2 -14 25
Both roots are imaginary.

5.

This program calculates the roots of a
quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 0 -14 28
The equation has the single root x = 2

6.

/tmp/AX1BKNaHpO.o
This program calculates the roots of a
        quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 2 0 25
Both roots are imaginary.

b) Following is the program for to output imaginary case as well :

#include <iostream>

#include <cmath>

using namespace std;

// this program solves for the roots of a quadratic equation

int main()

{

double a, b, c, disc, root1, root2;

cout << "This program calculates the roots of a\n";

cout << "\tquadratic equation of the form\n";

cout << "\t\t 2\n";

cout << "\t\tax + bx + c = 0\n\n";

cout << "Please enter values for a, b, and c: ";

cin >> a >> b >> c;

if ( a == 0.0 && b == 0.0)

cout << "The equation is degenerate and has no roots.\n";

else if (a == 0.0)

cout << "The equation has the single root x = " << -c/b << endl;

else {

disc = pow(b,2.0) - 4 * a * c;    // calculate discriminant

if (disc > 0.0) {

disc = sqrt(disc);

root1 = (-b + disc) / (2 * a);

root2 = (-b - disc) / (2 * a);

cout << "The two real roots are " << root1 << " and " << root2

<< endl;

}

else if (disc < 0.0)
{
double x,y;
x=-b/(2*a);
y= sqrt(-disc)/(2*a);
cout << "The two complex roots are "<<x<<"+"<<y<<"i"<<endl;
}

else

cout << "Both roots are equal to " << -b / (2 * a) << endl;

}

return 0;

}

Code demo :

Output :

Console copy-pasted output :

This program calculates the roots of a
quadratic equation of the form
                 2
                ax + bx + c = 0

Please enter values for a, b, and c: 2 -14 25
The two complex roots are 3.5+0.5i