Improved Quadratic Equation Solver

from __init__ import install_dependencies

await install_dependencies()

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

Source

# 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

Source

# 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

Source

# 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))