This topic is significant because it is the first example of pure algebraic manipulation!

The process is probably the first regular encounter with algebra in which an equation is not directly solved, but rather expressions are just manipulated into different forms to reveal information which needs further manipulation or interpretation

## complete the square

Strictly, the process of completing the square is the process of rewriting $x^2 + 2bx$ as $(x+b)^2 - b^2$. This strict process can be interpreted geometrically as shown in the figure.

The red and green regions have a total area of $x^2 + 2bx$. The entire square has an area of $(x+b)^2$. It follows then that the area of the red and green regions can be expressed as $x^2 + 2bx \equiv (x+b)^2 - b^2$.

Extending this geometric argument to the cases where $b$ is negative is a little less satisfactory, though of course entirely possible.

We can also extend this process to cover cases of the form $ax^2 + bx + c$, where $a\neq 1$

In the case where $a\neq 1$, we need to preform an extra couple of steps.

- Rewrite $ax^2 + bx + c$ as $a[x^2 + \frac{b}{a}x] + c$, introducing square brackets to separate out the $a$ term
- Complete the square inside the square bracket, to give $$a\left[\left(x-\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c$$
- Expand the square bracket to give $$a\left(x-\frac{b}{2a}\right)^2 -\frac{b^2 - 4ac}{4a}$$

You may notice that this expression is beginning to look startlingly similar to the quadratic formula, and indeed a few more manipulations would yield the quadratic formula.

We can now attempt to use the technique of completing the square on some expressions and look at the interpretation of the results.

## complete the square 1

The expression $x^2 + 12x$ can be rewritten as follows $$ \begin{align} x^2 + 12x &= (x+6)^2 - 6^2 \\ &= (x+6)^2 - 36 \end{align} $$

It follows that the equation of the curve can now be rewritten as $y = (x+6)^2 - 36$.

Now note that when we square a **real** number, the answer can never be negative, so $$(x+6)^2 \geq 0$$ The expression is only equal to $0$ when $x=-6$.

It follows that $y \geq 0-36$ and only equals $-36$ when $x = -6$. This is illustrated in the plot below.

## completing the square 2

By completing the square, rewrite the expression $x^2 + 4x - 8$ in the form $(x+a)^2 + b$, hence or otherwise determine the roots of the curve with equation $y = x^2 + 4x - 8$.

The expression $x^2 + 4x - 8$ can be rewritten as follows $$ \begin{align} x^2 + 4x - 8 &= (x+2)^2 - 2^2 - 8\\ &= (x+2)^2 - 4 - 8 \\ &= (x+2)^2 - 12 \end{align} $$

The expression given in the question does not factorise, so we shall proceed by rearranging the expression we have just derived. The roots of the curve occur when $y=0$

$$ \begin{align} 0 &= x^2 + 4x -8 \\ 0 &= (x+2)^2 -12\\ 12 &= (x+2)^2 \\ \pm\sqrt{12} &= x + 2 \\ x &= -2\pm \sqrt{12} \\ x &= -2\pm 2\sqrt{3} \end{align}$$

## completing the square 3

We can complete the square on this equation, despite the fact that it is a quartic!

$$ \begin{align} 4t^4 - 8t^2 - 15 &= 4\left[t^4 - 2t^2\right] - 15 \\ &= 4\left[(t^2-1)^2 - 1^2\right] - 15 \\ &= (t^2-1)^2 - 19 \end{align} $$The solutions of the equation can now be calculated $$ \begin{align} (t^2-1)^2 - 19 &= 0 \\ (t^2-1)^2 &= 19 \\ (t^2 - 1) &= \pm\sqrt{19} \\ t^2 &= 1 \pm \sqrt{19} \\ t &= \pm \sqrt{1\pm \sqrt{19}} \end{align}$$ Now note that $1-\sqrt{19}$ is negative and therefore doesn't have a **real** square root.

The solutions are therefore $$ t = \pm \sqrt{1 + \sqrt{19}} $$