# The Square-Complete Quadratic Expression

Updated: 2-Sep-2019

Posted: 25-Aug-2019

## 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

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.