Second-order polynomial equations with Polya's method

Much like a chef is tested with a basic omelette in a cooking interview to show essential kitchen skills, tackling second-order polynomial equations reflects similar foundational concepts in algebra. This task, while seemingly simple, highlights the necessity of grasping and utilizing core principles. Here, we’ll explore this concept using Polya’s problem-solving steps, offering a fresh perspective on a classic mathematical challenge.

It is well known that when confronted with second-order polynomial equations of the general form:

$$\begin{equation} ax^2 + bx + c = 0 \end{equation}$$

the presence of both quadratic and linear terms complicates the solution. The solution to this equation is well known, our added value for today is to address it using Polya’s problem-solving steps.

Understanding the Problem

Why is a quadratic equation, specifically one of the form above, challenging to solve? The difficulty arises from the inability to isolate $x$ by means of the basic algebraic operations. This complexity calls for a strategic approach to facilitate the solution process.

Simplifying the Problem: Monic Form

First, we transform our equation into a monic form to standardize the approach:

$$\begin{equation} x^2 + px + q = 0, \quad \text{where } p = \frac{b}{a} \text{ and } q = \frac{c}{a}. \end{equation}$$

This simplification reduces the problem to a more familiar format.

Now we ask one of Polya’s questions from this step: can we solve a simpler problem? Yes, the solution to the quadratic equation $y^2 = r$ can be found in a one-step operation, by taking the square root of $r$.

Let’s work in this direction.

Devising a Plan

To eliminate the linear term (also known as depressing the polynomial, we borrow this idea from the strategy to solve third-order polynomials), we propose a linear change of variables:

$$\begin{equation} x = y + k. \end{equation}$$

Our objective is to choose $k$ such that it cancels out the linear term in the transformed equation, thus reducing our original equation to one that is more straightforward.

The steps are:

1. Substitute the new variable into the original equation.
2. Simplify the equation by expanding and collecting like terms.
3. Choose $k$ such that the linear term is eliminated.
4. Solve for $y$.
5. Revert the change of variables to find $x$.

Executing the Plan

Substitution

$$\begin{equation} (y + k)^2 + p(y + k) + q = 0. \end{equation}$$

Simplification

$$\begin{align} y^2 + 2ky + k^2 + py + pk + q &= 0, \\ y^2 + (2k + p)y + (k^2 + pk + q) &= 0. \end{align}$$

Choosing $k$

We choose $k$ such that the coefficient of $y$ is zero:

$$\begin{equation} 2k + p = 0 \implies k = -\frac{p}{2}. \end{equation}$$

The other terms with $k$ are:

$$\begin{align} k^2 &= \frac{p^2}{4}, \\[3mm] pk &= -\frac{p^2}{2}, \\[3mm] k^2 + pk + q &= -\frac{p^2}{4} + q. \end{align}$$

Once the algebraic dust settles, the quadratic equation for $y$ is:

$$\begin{equation} y^2 = \frac{p^2}{4} - q, \end{equation}$$

and so the solutions are:

$$\begin{equation} y = \pm \sqrt{\frac{p^2}{4} - q}. \end{equation}$$

Before we revert the change of variables, we apply one more simplifcation step to the equation in $y$, to take the $1/4$ out of the square root:

$$\begin{equation} y = \pm \frac{1}{2} \sqrt{p^2 - 4q}. \end{equation}$$

Revert the Change of Variables

Finally, we revert the change of variables to find $x$:

$$\begin{equation} x = -\frac{p}{2} \pm \frac{1}{2} \sqrt{p^2 - 4q}. \end{equation}$$

We can further revert the transformation to the original variables $a$, $b$, and $c$:

$$\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \end{equation}$$

which is the well-known quadratic formula.

Looking Back

We aim to highlight the benefits of simplifying problems. Changing variables is a useful algebraic tool and a general approach to problem-solving. This technique shows the practicality of organized thought.

By examining second-order polynomial equations with this approach, we enhance our grasp of math problem-solving methods.