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

$x^2+xy$ $=$ $x^2+xy+0$

$=$ $x^2+xy+{\displaystyle \frac{y^2}{4}}-{\displaystyle \frac{y^2}{4}}$

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

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

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

  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

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

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

  $\ge$ $5,$

  since ${(2x+2)}^2\ge 0$. Here, equality holds if and only if $x=-1$. Thus $p(x)\ge 5$ for all $x\in ℝ$, 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].

(Anyone has an English reference?)

Title	completing the square
Canonical name	CompletingTheSquare
Date of creation	2013-03-22 13:36:27
Last modified on	2013-03-22 13:36:27
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	14
Author	mathcam (2727)
Entry type	Algorithm
Classification	msc 00A20
Related topic	SquareOfSum