completing the square

As always, we need our friends.

This is just a quick reference on how to solve quadratic equations using square completion. We'll use sympy and matplotlib to verify our results.

Completing the square is a basic method of solving a quadratic equation. In fact, the quadratic formula can be proven using the method of completing the square.

$(x + a)^2 = x^2 + 2ax + a^2$

We'll start with a basic quadratic equation by writing down a random one: $x^2 + 5x - 18 = 0$

Note that you can write down anything like $ax^2 + bx + c$ and just substitude $a$, $b$ and $c$ with some random numbers and it will work. You might run into negative roots but then you can just involve the imaginary number and solve it anyways (for example, see the second example below).

What we are aiming for is to write it in a square form such as $(x + a)^2 = y$ so we can actually solve it by saying $x + a = \sqrt{y}$.

Let's get back to our equation $x^2 + 5x - 18 = 0$. We'll start by adding $18$ to each side:

$$ \begin{align} x^2 + 5x - 18 & = 0 \\ x^2 + 5x & = 18 \end{align} $$

Just as a reminder:

$(x + a)^2 = (x + a) \times (x + a) = x \times x + a \times x + a \times x + a \times a = x^2 + 2ax + a^2$

In this case $2ax = 5x$ so in order to figure out what $a$ is we need to divide both sides by $2x$. This gives us $a = \frac{5}{2} = 2\frac{1}{2}$. We need to add $a^2 = {2\frac{1}{2}}^2 = \frac{25}{4} = 6\frac{1}{4}$ to complete the square and the rules of algebra tell us that we need to add it to each side:

$$ \begin{align} x^2 + 5x & = 18 \\ x^2 + 5x + 6\frac{1}{4} & = 18 + 6\frac{1}{4} \\ x^2 + 5x + 6\frac{1}{4} & = 24\frac{1}{4} \end{align} $$

Now we have a constant number on the right-hand side. It's a rational but that's no problem. At least its positive so we don't need $i$ for this. More importantly though, the left-hand side is in a $x^2 + 2ax + a^2$ configuration. That means we can now rewrite it as $(x + a)^2$.

$$ \begin{align} x^2 + 5x + 6\frac{1}{4} & = 24\frac{1}{4} \\ (x - 2\frac{1}{2})^2 & = 24\frac{1}{4} \end{align} $$

So now that we have a square expression on the left and a constant on the right we can start to formulate an answer:

$$ \begin{align} (x + 2\frac{1}{2}) = \pm\sqrt{24\frac{1}{4}} \implies x & = -2\frac{1}{2} - \sqrt{24\frac{1}{4}} \cup -2\frac{1}{2} + \sqrt{24\frac{1}{4}} \\ & = -2\frac{1}{2} - \frac{1}{2}\sqrt{97} \cup -2\frac{1}{2} + \frac{1}{2}\sqrt{97} \end{align} $$

We can check this using sympy.

Bcause we can and because it's kinda fun and easy to plot we'll try to visualize the solution as well.

imaginary example

So let's try another one. Again, we'll just jot down some random quadratic equation: $3x^2 - 3x + 5 = 0$. We want to get rid of that $3$ at the beginning so we'll start by scaling down the whole by a factor of $3$.

$\frac{3x^2 - 3x + 5}{3} = 0 \implies x^2 - x + \frac{5}{3} = 0 $

Now that we have something a bit easier to work with, the next thing we want is to write the left hand side as $x^2 + 2ax + a^2$.

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$$ \begin{align} x^2 - x + \frac{5}{3} & = 0 \\ x^2 - x & = -\frac{5}{3} & \text{subtract $\frac{5}{3}$} \\ x^2 - x + \frac{1}{4} & = -\frac{5}{3} + \frac{1}{4} & \text{add $a^2$ with $a = \frac{1}{2}$} \\ x^2 - x + \frac{1}{4} & = -\frac{17}{12} & \text{simplify} \\ (x - \frac{1}{2})^2 & = -\frac{17}{12} & \text{write as square} \\ x - \frac{1}{2} & = \pm \sqrt{-\frac{17}{12}} & \text{unsquare left-hand side} \\ x & = \frac{1}{2} \pm \sqrt{\frac{17}{12}}i & \text{remove negative root using $i$} \\ x & = \frac{1}{2} \pm \sqrt{51}\frac{i}{6} & \text{simplify} \end{align} $$

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Let's check if sympy agrees with our answer.

Looks like we got it. To be honest, I kinda cheated on the last simplify step by using sympy to simplify the square root of the rational. Let's make sure our initial answer was correct.