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 valueof 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 andonly 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 techniqueto 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?)