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'completing the square'
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| Title of object: |
completing the square |
| Canonical Name: |
CompletingTheSquare |
| Type: |
Algorithm |
| Created on: |
2003-05-02 07:14:40 |
| Modified on: |
2003-06-23 09:00:04 |
| Classification: |
msc:00A20 |
Revision comment (for changes between this and next version):
| Changes for correction #6054 ('portability'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here |
Content:
\newcommand{\sR}[0]{\mathbb{R}}
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
\begin{eqnarray*}
x^2+xy &=& x^2+xy+ 0\\
&=& x^2+xy+ \frac{y^2}{4}-\frac{y^2}{4}\\
&=& (x+\frac{y}{2})^2-\frac{y^2}{4}.
\end{eqnarray*}
This manipulation is called \emph{completing the square} \cite{adams} in
$x^2+xy$, or completing the square $x^2$.
Replacing $y$ by $-y$, we also have
$$x^2-xy = (x-\frac{y}{2})^2-\frac{y^2}{4}.$$
Here are some applications of this method:
\begin{itemize}
\item
\PMlinkname{Derivation of the solution formula to the quadratic equation}{DerivationOfQuadraticFormula}.
\item Completing the square can also be used to find the extremal value
of a quadratic polynomial \cite{thompson} without calculus.
Let us illustrate this for the polynomial $p(x)=4x^2+8x+9$.
Completing the square yields
\begin{eqnarray*}
p(x) &=& (2x+2)^2-4 +9 \\
&=& (2x+2)^2+5 \\
&\ge & 5,
\end{eqnarray*}
since $(2x+2)^2\ge 0$. Here, equality holds if and
only if $x=-1$.
Thus $p(x)\ge 5$ for all $x\in \sR$, and $p(x)=5$ if and only if
$x=-1$.
It follows that $p(x)$ has a global minimum at $x=-1$, where $p(-1)=5$.
\item Completing the square can also be used as an integration technique
to integrate, say $\frac{1}{4x^2+8x+9}$ \cite{adams}.
\end{itemize}
\begin{thebibliography}{9}
\bibitem {adams} R. Adams, \emph{Calculus, a complete course},
Addison-Wesley Publishers Ltd, 3rd ed.
\bibitem {thompson}
\emph{Matematik Lexikon} (in Swedish),
J. Thompson, T. Martinsson, Wahlstr\"om \& Widstrand, 1991.
\end{thebibliography}
(Anyone has an English reference?) |
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