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4. Improved Quadratic Equation Solver

In this notebook, we will improve the quadratic equation solver in the previous lab using conditional executions.
First of all, run the following to setup the environment.

%reset -f
from ipywidgets import interact
import math

4.1. Zero Discriminant

Recall that the quadratic equation is

\[ ax^2+bx+c=0 \]

where \(a\), \(b\), and \(c\) are real-valued coefficients, and \(x\) is the unknown variable. The roots are normally given by

\[ \frac{-b-\sqrt{b^2-4ac}}{2a}, \frac{-b+\sqrt{b^2-4ac}}{2a}. \]

The roots are the same (repeated) when the discriminant \(b^2-4ac\) is zero.

Exercise Assign to roots only one root when the discriminant is zero. E.g., if \((a,b,c)=(1,-2,1)\), then roots should be assigned the value 1.0 instead of 1.0, 1.0. If there are two roots, give them in the order of the above formula.

Hint: Use the if statement.

Hint: The following is a solution template with some missing code. You are NOT required to follow the template.

def get_roots(a, b, c):
    d = b**2 - 4 * a * c    # discriminant
    if math.isclose(d, 0):
        roots = __________  # repeated root
    else:
        d **= 0.5
        roots = __________________________________
    return roots

def get_roots(a, b, c):
    d = b**2 - 4 * a * c    # discriminant
    if math.isclose(d, 0):
    # YOUR CODE HERE
    raise NotImplementedError()
    return roots

# tests
def test_get_roots(roots, a, b, c):
    roots_ = get_roots(a, b, c)
    if roots is None:
        correct = roots_ is None
    elif isinstance(roots, float):
        correct = isinstance(roots_, float) and math.isclose(roots, roots_)
    else:
        correct = isinstance(roots_, tuple) and len(roots_) == 2 and all([
            math.isclose(root, roots_) for root, roots_ in zip(roots, roots_)
        ])
    if not correct:
        print(f'With (a, b, c)={a,b,c}, roots should be {roots} not {roots_}.')
    assert correct

test_get_roots((-1.0, 0.0), 1, 1, 0)
test_get_roots(0.0, 1, 0, 0)

Exercise Why use math.isclose(d,0) instead of d == 0?

YOUR ANSWER HERE

4.2. Linear Equation

If \(a=0\), the earlier formula for the roots are invalid due to division by zero. Nevertheless, the equation remains valid:

\[ bx + c=0. \]

Exercise Improve the function get_roots to return the root \(-\frac{c}{b}\) if \(a=0\).

Hint: Solution template:

def get_roots(a, b, c):
    d = b**2 - 4 * a * c    # discriminant
    if __________________:
        roots = ______
    elif math.isclose(d, 0):
        roots = __________  # repeated root
    else:
        d **= 0.5
        roots = __________________________________
    return roots

def get_roots(a, b, c):
    d = b**2 - 4 * a * c
    # YOUR CODE HERE
    raise NotImplementedError()
    return roots

# tests
def test_get_roots(roots, a, b, c):
    roots_ = get_roots(a, b, c)
    if roots is None:
        correct = roots_ is None
    elif isinstance(roots, float):
        correct = isinstance(roots_, float) and math.isclose(roots, roots_)
    else:
        correct = isinstance(roots_, tuple) and len(roots_) == 2 and all([
            math.isclose(root, roots_) for root, roots_ in zip(roots, roots_)
        ])
    if not correct:
        print(f'With (a, b, c)={a,b,c}, roots should be {roots} not {roots_}.')
    assert correct


test_get_roots((-1.0, -0.0), 1, 1, 0)
test_get_roots(0.0, 1, 0, 0)
test_get_roots(0.5, 0, -2, 1)

4.3. Degenerate Cases

What if \(a=b=0\)? In that case, the equation becomes

\[ c = 0 \]

which is always satisfied if \(c=0\), but never satisfied if \(c\neq 0\).

Exercise Improve the function get_roots to return root(s) under all cases:

  • If \(a=0\) and \(b\neq 0\), assign roots to the single root \(-\frac{c}{b}\).

  • If \(a=b=0\) and \(c\neq 0\), assign roots to None.
    Note that None is an object, not a string.

  • If \(a=b=c=0\), there are infinitely many roots. Assign to roots the tuple -float('inf'), float('inf').
    Note that float('inf') converts the string 'inf' to a floating point value that represents \(\infty\).

Hint: Use nested if statements such as the followings (with the blanks filled in properly):

def get_roots(a, b, c):
    d = b**2 - 4 * a * c
    if __________________:
        if __________________:
            if __________________:
                roots = -float('inf'), float('inf')
            else:
                roots = None
        else:
            ______________
    elif math.isclose(d, 0):
        roots = __________  # repeated root
    else:
        d **= 0.5
        roots = __________________________________
    return roots

def get_roots(a, b, c):
    d = b**2 - 4 * a * c
    # YOUR CODE HERE
    raise NotImplementedError()
    return roots

# tests
def test_get_roots(roots, a, b, c):
    roots_ = get_roots(a, b, c)
    if roots is None:
        correct = roots_ is None
    elif isinstance(roots, float):
        correct = isinstance(roots_, float) and math.isclose(roots, roots_)
    else:
        correct = isinstance(roots_, tuple) and len(roots_) == 2 and all([
            math.isclose(root, roots_) for root, roots_ in zip(roots, roots_)
        ])
    if not correct:
        print(f'With (a, b, c)={a,b,c}, roots should be {roots} not {roots_}.')
    assert correct


test_get_roots((-1.0, 0.0), 1, 1, 0)
test_get_roots(0.0, 1, 0, 0)
test_get_roots((-float('inf'), float('inf')), 0, 0, 0)
test_get_roots(None, 0, 0, 1)
test_get_roots(0.5, 0, -2, 1)
test_get_roots(1.0, 1, -2, 1)

4.4. Run the calculator

After you have complete the exercises, you can run your robust solver below:

# quadratic equations solver
@interact(a=(-10,10,1),b=(-10,10,1),c=(-10,10,1))
def quadratic_equation_solver(a=1,b=2,c=1):
    print('Root(s):',get_roots(a,b,c))
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