## Background

Consider the expression

$(x + h)^2$

Expanding it by the Distribution Axiom

\begin{aligned} (x + h)^2 &\equiv (x + h)(x + h) \\ &\equiv (x + h)x + (x + h)h \\ &\equiv x^2 + xh + xh + h^2 \\ &\equiv x^2 + 2xh + h^2 \\ \end{aligned}

The final result is

$\tag{1}(x + h)^2 \equiv x^2 + 2xh + h^2$

which when rearranged becomes the identity

$\tag{2} \boxed{x^2 + 2xh \equiv (x + h)^2 -h^2}$

The importance of the above identity is it shows a quadratic expression (on left side of identity) that has two terms, $$x^2$$ and $$x$$, could be transformed into one that has only a single unknown $$x$$ term (on right side).

Identity $$(2)$$ should be enough for us to proceed to look at the application of it; but many of us overlook the meaning behind symbols such as $$x$$'s and $$h$$'s in Mathematics. To overcome this, I am going to deliberately state identity $$(2)$$ using more 'graphic' symbols hoping the patterns in the identity present themselves clearer.

$\tag{3} \boxed{\spades^2 + {\color{red}2}\thinspace \spades\hearts \equiv (\spades + \hearts)^2 -\hearts^2}$

The above identity $$(3)$$ tells us, an expression-pattern or shape as on the left side of the identify $$(\spades^2 + 2\thinspace \spades\hearts)$$, is completely equivalent to the expression on the right side of the identity $$((\spades + \hearts)^2 -\hearts^2)$$, for any two numbers $$\spades$$ and $$\hearts$$.

Note the number $$\color{red}2$$ is there in the identify as a natural consequence of a series of valid mathematical operations. Look at the derivation of the identity again if you're suspectful why the number $$\color{red}2$$ should be there.

All these are fine but what is a square-complete quadratic expression? It is simply any expression of the shape

$\spades^2 + 2\thinspace \spades\hearts + \hearts^2$

The above expression is deemed square-complete because it could be 'repackaged' (factorised) into

$(\spades + \hearts)^2$

See identity $$(1)$$ - but replace $$x$$ and $$h$$ with $$\spades$$ and $$\hearts$$ respectively. We could also obtain $$(2)$$ using the concept of areas. This approach might appeal more to people who like to think in terms of pictures and diagrams.

Notice in the above diagram, Figure-1, there is a square block of $$x^2$$ units², with two rectangle blocks of $$xd$$ units² each. The total area of these blocks is

$x^2 + 2xd$

Now, we could also view the above square and rectangle blocks as part of a larger square block of $$(x+d)^2$$ units² - see Figure-2, below.

This enables us to obtain the same area as before with the expression

$(x+d)^2 - d^2$

Because both expressions give the same area, they're equivalent, giving us the same result as $$(2)$$

$x^2 + 2xd \equiv (x+d)^2 - d^2$

## Application

Let's see how the result obtained in the theory could be used with actual numbers. Consider

$x^2 - 5x + 3$

Do you see the shape $$\spades^2 + 2\thinspace \spades\hearts$$ in the above expression? If you don't, with a bit of manipulation

\begin{aligned} x^2 - 5x {\color{silver} + 3} &= \underbrace{x^2 + 2x\left(-\frac{5}{2}\right)}_{\spades^2 + 2\thinspace \spades\hearts} {\color{silver} + 3} \end{aligned}

Now we could move on

\begin{aligned} {x^2 - 5x} {\color{silver}+ 3} &= {x^2 + 2x\left(-\frac{5}{2}\right)} {\color{silver}+ 3} \\ &= \underbrace{\left(x + \left[-\frac{5}{2}\right]\right)^2 - \left[-\frac{5}{2}\right]^2 }_{(\spades + \hearts)^2 - \hearts^2} {\color{silver}+ 3} \end{aligned}

Further tidying

\begin{aligned} x^2 - 5x + 3 &= \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + 3 \\ &= \left(x - \frac{5}{2}\right)^2 - \frac{13}{4} \end{aligned}

In conclusion

$x^2 - 5x + 3 = \left(x - \frac{5}{2}\right)^2 - \frac{13}{4}$

Just what is the fuss about presenting quadratic expressions in terms of square-complete phrases? Suppose we have the following equation to solve

$x^2 - 5x + 3 = 0$

We're required to find the values for the unknown $$x$$ that satisfies the equation. We will use the square-complete form to achieve this

\begin{aligned} x^2 - 5x + 3 &= 0 \\ \left(x - \frac{5}{2}\right)^2 - \frac{13}{4} &= 0 \\ \left(x - \frac{5}{2}\right)^2 &= \frac{13}{4} \\ x - \frac{5}{2} &= \pm{\frac{\sqrt{13}}{2}} \\ x &= \frac{5 \pm \sqrt{13}}{2} \quad \blacksquare \end{aligned}

As you can see, the square-complete technique converts a quadratic expression that has $$x^2$$ and $$x$$ terms into a form that has only one $$x$$ term, enabling us to solve the equation.