completing the square

Let us consider the expression $x^{2}+xy$ , where $x$ and $y$ are real (or complex) numbers. Using the formula

(x+y)^{2}=x^{2}+2xy+y^{2}

we can write

$\displaystyle x^{2}+xy$	$\displaystyle=$	$\displaystyle x^{2}+xy+0$
	$\displaystyle=$	$\displaystyle x^{2}+xy+\frac{y^{2}}{4}-\frac{y^{2}}{4}$
	$\displaystyle=$	$\displaystyle\left(x+\frac{y}{2}\right)^{2}-\frac{y^{2}}{4}.$

This manipulation is called completing the square [1] in $x^{2}+xy$ , or completing the square $x^{2}$ .

Replacing $y$ by $-y$ , we also have

x^{2}-xy=\left(x-\frac{y}{2}\right)^{2}-\frac{y^{2}}{4}.

Here are some applications of this method:

•

http://planetmath.org/DerivationOfQuadraticFormulaDerivation of the solution formula to the quadratic equation.
•

Putting the general equation of a circle, ellipse, or hyperbola into standard form, e.g. the circle

$\displaystyle x^{2}+y^{2}+2x+4y=5\Rightarrow(x+1)^{2}+(y+2)^{2}=10,$

from which it is frequently easier to read off important information (the center, radius, etc.)
•

Completing the square can also be used to find the extremal value of a quadratic polynomial [2] without calculus. Let us illustrate this for the polynomial $p(x)=4x^{2}+8x+9$ . Completing the square yields

$\displaystyle p(x)$ $\displaystyle=$ $\displaystyle(2x+2)^{2}-4+9$

$\displaystyle=$ $\displaystyle(2x+2)^{2}+5$

$\displaystyle\geq$ $\displaystyle 5,$

since $(2x+2)^{2}\geq 0$ . Here, equality holds if and only if $x=-1$ . Thus $p(x)\geq 5$ for all $x\in\mathbb{R}$ , and $p(x)=5$ if and only if $x=-1$ . It follows that $p(x)$ has a global minimum at $x=-1$ , where $p(-1)=5$ .
•

Completing the square can also be used as an integration technique to integrate, for example the function $\displaystyle\frac{1}{4x^{2}+8x+9}$ [1].

References

1 R. Adams, Calculus, a complete course, Addison-Wesley Publishers Ltd, 3rd ed.
2 Matematiklexikon (in Swedish), J. Thompson, T. Martinsson, Wahlström & Widstrand, 1991.

(Anyone has an English reference?)