Quadrados perfeitos

So, what exactly is a "perfect square"?

A perfect square is the product of two identical terms multiplied together. Here are some examples:

$$\begin{align*} 1 \cdot 1 = 1\\ 2 \cdot 2 = 4\\ 3 \cdot 3 = 9 \\ 4 \cdot 4 = 16 \\ 5 \cdot 5 = 25 \end{align*}$$

Notice how we said "two identical terms multiplied together" instead of "two identical numbers"? This is because terms with variables can also be considered perfect squares.

The same principle holds true: if it's two identical terms being multiplied together, the product is a perfect square.

So, these terms (highlighted in purple) can also be considered perfect squares:

$$\begin{align*} x \cdot x = \color{Purple}{x^2}\\ 2x \cdot 2x = \color{Purple}{4x^2}\\ (x-1) \cdot (x-1) = \color{Purple}{(x-1)^2}\\ (x+4) \cdot (x+4) = \color{Purple}{(x+4)^2} \end{align*}$$

What do perfect squares have to do with quadratic equations?

You can spot a perfect square by looking at the first and last terms, $ax^2$ and $c$, in a quadratic equation $ax^2+bx+c=0$. If both of these terms are perfect squares, then we can sometimes use this information to quickly factor the left hand side our equation.

Consider the equation $x^2+10x+25=0$. Let's take a look at the left hand expression:

Both of these terms are perfect squares. In fact, the quadratic expression itself can be written as a perfect square: as two identical factors multiplied together. You can FOIL to check that $(x+5)(x+5) = x^2+10x+25$.

This tells us that the only zero, or solution, to $(x+5)^2 = 0$, is $x=-5$.

What Are They?

A perfect square quadratic equation is just a quadratic equation that can be factored into the product of two of the same binomial. For example, $(2x+4)^2$ is a perfect square quadratic expression but factored. When we expand it, it is $4x^2+16x+16$

How do You Solve It?

Step 1: Like any other quadratic equation we first factor it. To do this we only have to look at the first and last term to see if they're perfect squares or not. For example, when you factor $$9x^2+72x+144=0$$ we can see $9x^2$ and $144$ are both squares, so the factorization is $(3x+12)(3x+12)=0$, which can be simplified to $(3x+12)^2=0$.

Step 2: Now to solving! It is just like normal, but instead of two roots there's only one because both of the terms in the factorization are the same. So our answer is $$x=-4$$

Step 3: To check if we factored right, all we have to do is expand it. $(3x+12)(3x+12)$ is $9x^2+72x+144$.

We find this by: $3x\cdot3x+3x\cdot12+12\cdot3x+12\cdot12$ = $9x^2+72x+144$

Yay we solved it!

There's also a trick for expanding squares like that. There's a formula and it states, $(a+b)^2$ expanded is $a^2+2ab+b^2$.

We say an integer is a perfect square if it is equal to the product of two of the same integer. For example $4$ is a perfect square because $2 \cdot 2 = 4$ and $169$ is a perfect square because $13 \cdot 13 = 169$ but $171$ is not.

Similarly, a perfect square expression is an expression that is the product of the same expression. For example

$$(x+1) \cdot (x+1) = x^2 + 2x + 1$$

Therefore, $x^2+2x+1$ is a perfect square expression. This was pretty easy to determine because I showed you the decomposition. What if I didn't give you the decomposition?

As you probably found out, this was substantially harder. In order to do this more efficiently, we need to better understand the form of a perfect square expressions.

Note that for this section, we are only concerned about expressions that are the square of linear expressions. That is, the square of linear terms of the form $ax+b$.

Just to be thurough, in order to multiply this expression out notice

$$(ax+b) \cdot (ax+b) = ax \cdot ax + ax \cdot b + b \cdot ax + b \cdot b$$ $$= a^2x^2 + 2abx + b^2$$

Just as a quick reminder, $|q|$ means the absolute value of $q$. The absolute value of $q$ is the distance away from $0$; for instance $|1| = 1$ and $|-1| = 1$ because both $1$ and $-1$ are $1$ away from $0$. Another way to think about this is that it gets rid of the negative sign if it exists.

In particular, notice that if $p = a^2$ and $r = b^2$ then

$$|q| = 2ab = 2 \cdot \sqrt{a^2 \cdot b^2} = 2 \cdot \sqrt{pr}$$.

(Why is the absolute value necessary)?

This turns out to be a sufficient condition to check if something is a square expression; this means that if $\sqrt{pr} = 2|q|$ then $px^2 + qx+r$ (think about why this is true!)

For instance, to check if $16x^2 + 16x + 4$ is a quadratic expression we notice that

$$2 \cdot \sqrt{16 \cdot 4} = 2 \cdot 8 = 16$$

which works! Now a step further would be considering how to factor this. Again, we use the idea that $p=a^2, q = 2ab, r=b^2$. Then, notice that

$$\begin{align*} a^2x^2 + 2abx + b^2 = &(ax+b)^2\\ &\implies px^2+qx+r = (\sqrt{p}x + \sqrt{r})^2 \end{align*}$$

Therefore

$$16x^2+16x+4 = (4x+2)^2$$