Improved Quadratic Equation Solver

import math
import numpy as np
from ipywidgets import interact

In this notebook, we will improve the quadratic equation solver in the previous lab using conditional executions.

Discriminant¶

Recall that a quadratic equation is

ax^2+bx+c=0

(1)

where $a$ , $b$ , and $c$ are real-valued coefficients, and $x$ is the unknown variable.

The discriminant Δ discriminates the roots

\frac{-b\pm \sqrt{\Delta}}{2a}

(3)

of a quadratic equation:

For example, if $(a,b,c)=(1,-2,1)$ , then

\Delta = (-2)^2 - 4(1\cdot 1) = 0

(5)

and so the repeated root is

- \frac{b}{2a} = \frac{2}2 = 1.

(6)

$(x-1)^2 = x^2 - 2x + 1$ as expected.

def iszero(v, a, b, c, *, rel_tol=1e-9):
    # YOUR CODE HERE
    raise NotImplementedError

# tests
assert not iszero(1e-8, 0, 2, 1)
assert iszero(1e-8, 1, 0, 1, rel_tol=1e-7)
assert not iszero(5e-9, 1, 2, 0)

# hidden tests

Exercise 2

Complete the following function by assigning roots only one root when the discriminant is zero using the relative tolerance test in Exercise 1. E.g., if (a, b, c) == (1, -2, 1), then roots should be assigned the value 1.0 instead of 1.0, 1.0.

Hint

You may use the if statement as follows:

def get_roots(a, b, c, *, rel_tol=1e-9):
    d = b**2 - 4 * a * c    # discriminant
    if iszero(d, a, b, c, rel_tol=rel_tol):
        roots = ...  # repeated root
    else:
        d **= 0.5
        roots = ...
    return root

Replace ... in the above code cell by your code. If you have a better implementation, you need not follow the solution template. E.g., you can write your solution entirely using boolean operations without any if statement.

def get_roots(a, b, c, *, rel_tol=1e-9):
    d = b**2 - 4 * a * c    # discriminant
    if iszero(d, a, b, c, rel_tol=rel_tol):
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()

YOUR ANSWER HERE

Linear equation¶

YOUR ANSWER HERE

Nevertheless, the quadratic equation remains valid:

bx + c=0,

(8)

the root of which is $x = -\frac{c}b$ .

Exercise 5

Improve the function get_roots to return the root $-\frac{c}{b}$ when $a=0$ .

Hint

Solution template:

def get_roots(a, b, c, *, rel_tol=1e-9):
    d = b**2 - 4 * a * c    # discriminant
    if ...:
        roots = ...
    elif iszero(d, a, b, c, rel_tol=rel_tol):
        roots = ...  # repeated root
    else:
        d **= 0.5
        roots = ...
    return roots

def get_roots(a, b, c, *, rel_tol=1e-9):
    d = b**2 - 4 * a * c
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()
assert np.isclose(get_roots(1, -2, 1), 1.0).all()
assert np.isclose(get_roots(0, -2, 1), 0.5).all()

Degenerate cases¶

What if $a=b=0$ ? In this case, the equation becomes

c = 0

(9)

which is always satisfied if $c=0$ but never satisfied if $c\neq 0$ .

def get_roots(a, b, c, *, rel_tol=1e-9):
    d = b**2 - 4 * a * c
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()
assert np.isclose(get_roots(1, -2, 1), 1.0).all()
assert np.isclose(get_roots(0, -2, 1), 0.5).all()
assert np.isclose(get_roots(0, 0, 0), (-float('inf'), float('inf'))).all()
assert get_roots(0, 0, 1) is None

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))