%% This document created by Scientific Notebook (R) Version 3.0


\documentclass[12pt,thmsa]{article}
\usepackage{amssymb}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{sw20jart}

%TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart}

%TCIDATA{<META NAME="ViewSettings" CONTENT="23">}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="Title" CONTENT="Other Equations">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sun Mar 23 14:45:23 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=Reference.cst}
%TCIDATA{PageSetup=72,72,72,72,0}
%TCIDATA{<LINK REL="contents" HREF="..\..\Reference\ReferenceIndex.tex#Reference Index">}
%TCIDATA{<LINK REL="next" HREF="Al02_06.tex">}
%TCIDATA{<LINK REL="parent" HREF="Al_contents.tex">}
%TCIDATA{<LINK REL="previous" HREF="Al02_04.tex">}
%TCIDATA{AllPages=
%F=36,\PARA{038<p type="texpara" tag="Body Text" >\hfill \thepage}
%}


\input{tcilatex}
\begin{document}


\section{Other Equations\label{Equations Other than Linear or Quadratic}}

The principles used to solve linear and quadratic equations can be used to
solve other types of equations, including some involving higher powers of
variables or radicals. For example, if you take an equation $A=B$ and
perform the \emph{same operation} on both sides, you still have a valid
equation (so, for instance, if $A$ and $B$ represent equal numbers, $A^{n}$
and $B^{n}$ represent equal numbers as well).

\subsection{Solving Equations Involving $n$th Roots\label%
{Solving Equations Involving nth Roots}}

\begin{itemize}
\item[Example]  Find all real solutions of 
\[
x^{3}=8
\]

\item[Solution]  Since every real number has exactly one real cube root, you
can solve this by taking the cube root of each side: 
\begin{eqnarray*}
\sqrt[3]{x^{3}} &=&\sqrt[3]{8} \\
x &=&2
\end{eqnarray*}

\item[Example]  Find all real solutions of 
\[
16x^{4}=81
\]

\item[Solution]  Here you must remember that if $n$ is even, every positive
real number has \emph{two} real $n$th roots, a positive one and a negative
one. So: 
\begin{eqnarray*}
x^{4} &=&\frac{81}{16}\text{\qquad Divide by }16 \\
\sqrt[4]{x^{4}} &=&\sqrt[4]{\frac{81}{16}}\text{\quad Take fourth roots} \\
x &=&\pm \frac{\sqrt[4]{81}}{\sqrt[4]{16}} \\
&=&\pm \frac{3}{2}
\end{eqnarray*}
\end{itemize}

\subsection{Taking Care of Extraneous Solutions\label%
{Extraneous Solutions of Equations}}

As you use the principal of performing the same operation on both sides of
an equation to simplify it, you may find extraneous solutions. This happens
because performing some operations on each side of an equation doesn't
necessarily result in an \hyperref{equivalent expression}{}{}{%
Al02_00.tex#Equivalent expressions}. Even if $A$ and $B$ aren't equal,
multiplying the false statement $A=B$ on both sides by $0$ results in the
true statement $0=0$. Another way to introduce extraneous solutions is to
raise each side of an equation to an even power. Even though $1=-1$ is
false, squaring both sides results in the true statement $1=1$. This means
that after you have performed such manipulations on an equation in order to
simplify it, you may be solving an equation which is true in some cases when
the original was false! In general, the best way of ensuring that solutions
are valid is simply by checking them in the original equation.

\begin{itemize}
\item[Example]  Find all solutions of 
\[
x+3=\frac{2x^{2}-7x+3}{3-x}
\]

\item[Solution]  Multiply both sides of the equation by $x-3$ to clear the
denominator: 
\begin{eqnarray*}
\left( x+3\right) \left( x-3\right)  &=&\frac{2x^{2}-7x+3}{-1} \\
x^{2}-9 &=&-2x^{2}+7x-3 \\
3x^{2}-7x-6 &=&0 \\
\left( 3x+2\right) \left( x-3\right)  &=&0 \\
x &=&-\frac{2}{3}\text{ or }x=3
\end{eqnarray*}
However, if $x=3$, is the original equation still satisfied? 
\begin{eqnarray*}
3+3 &=&\frac{2\left( 3\right) ^{2}-7\left( 3\right) +3}{3-3} \\
6 &=&\frac{0}{0}
\end{eqnarray*}
This is \emph{false}, since the right hand side isn't even a valid algebraic
expression. So $x=-\frac{2}{3}$ is the only solution. Note than if $x=3$,
then our initial step to clear the denominator amounted to multiplying each
side by $0$ - one of our error-prone cases mentioned above! But it slipped
in almost without our noticing it...

\item[Example]  Find all solutions of 
\[
x=1-\sqrt{2-\frac{x}{2}}
\]

\item[Solution]  First, isolate the radical by subtracting $1$ from each
side: 
\[
x-1=-\sqrt{2-\frac{x}{2}}
\]
Now square each side and simplify: 
\begin{eqnarray*}
\left( x-1\right) ^{2} &=&2-\frac{x}{2} \\
x^{2}-2x+1 &=&2-\frac{x}{2} \\
2x^{2}-4x+2 &=&4-x \\
2x^{2}-3x-2 &=&0 \\
\left( 2x+1\right) \left( x-2\right)  &=&0 \\
2x+1 &=&0\text{ or }x-2=0 \\
x &=&-\frac{1}{2}\text{ or }x=2
\end{eqnarray*}
But now check each of these in the original. If $x=-\frac{1}{2}$%
\begin{eqnarray*}
-\frac{1}{2} &=&1-\sqrt{2-\frac{-\frac{1}{2}}{2}}? \\
-\frac{1}{2} &=&1-\sqrt{2+\frac{1}{4}}? \\
-\frac{1}{2} &=&1-\sqrt{\frac{9}{4}}? \\
-\frac{1}{2} &=&1-\frac{3}{2}\checkmark 
\end{eqnarray*}
But if you try $x=2$%
\begin{eqnarray*}
2 &=&1-\sqrt{2-\frac{2}{2}}? \\
2 &=&1-\sqrt{1}? \\
2 &\neq &0
\end{eqnarray*}
So $x=-\frac{1}{2}$ is the only valid solution.
\end{itemize}

\subsection{Equations of Quadratic Type}

Some equations may be regarded as quadratic equations by making a suitable
substitution.

\begin{itemize}
\item[Example]  Find the real solutions of 
\[
x^{4}-2x^{2}-2=0
\]

\item[Solution]  Notice that all the terms appearing in this equation can be
written in terms of $x^{2}$, since $x^{4}=\left( x^{2}\right) ^{2}$. So use
a substitution, replacing $x^{2}$ by $w$: 
\[
w^{2}-2w-2=0
\]
Now use the \hyperref{quadratic formula}{}{}{Al02_03.tex#Quadratic Formula}
to find solutions for $w$: 
\begin{eqnarray*}
w &=&\frac{-\left( -2\right) \pm \sqrt{\left( -2\right) ^{2}-4\left(
1\right) \left( -2\right) }}{2\left( 1\right) } \\
&=&\frac{2\pm \sqrt{4+8}}{2} \\
&=&\frac{2\pm \sqrt{12}}{2} \\
&=&\frac{2\pm 2\sqrt{3}}{2} \\
&=&1\pm \sqrt{3}
\end{eqnarray*}
Now you know the possible values for $w$, which equals $x^{2}$; so: 
\begin{eqnarray*}
x^{2} &=&1\pm \sqrt{3} \\
x &=&\pm \sqrt{1\pm \sqrt{3}}
\end{eqnarray*}
This gives us (potentially) four solutions; but since you're looking only at
real solutions you must ensure that the quantity under the radical is
nonnegative. $1+\sqrt{3}$ is clearly a positive number, but, since $\sqrt{3}%
>1$, $1-\sqrt{3}$ is negative. So our final solutions are 
\[
x=\pm \sqrt{1+\sqrt{3}}
\]

\item[Example]  Find all real solutions of 
\[
x^{6}-2x^{3}=-2-\left( x^{6}-2x^{3}\right) 
\]

\item[Solution]  First simplify: 
\begin{eqnarray*}
x^{6}-2x^{3} &=&-2-x^{6}+2x^{3} \\
2x^{6}-4x^{3}+2 &=&0
\end{eqnarray*}
Now, recognizing that all the terms can be expressed in terms of $x^{3}$
(since $x^{6}=\left( x^{3}\right) ^{2}$), let $v=x^{3}$ and rewrite the
equation as: 
\begin{eqnarray*}
2v^{2}-4v+2 &=&0 \\
v^{2}-2v+1 &=&0 \\
\left( v-1\right) ^{2} &=&0 \\
v &=&1
\end{eqnarray*}
Since $v=x^{3}$, this means: 
\begin{eqnarray*}
x^{3} &=&1 \\
x &=&1\text{ (taking cube roots)}
\end{eqnarray*}
\end{itemize}

\subsection{Solutions by Factoring\label%
{Solutions to General Equations by Factoring}}

\subsubsection{Factoring using special products}

\begin{itemize}
\item[Example]  Find all roots of 
\[
x^{6}-1=0
\]

\item[Solution]  Noticing that $x^{6}=\left( x^{3}\right) ^{2}$, use 
\hyperref{difference of squares factoring}{}{}{Al01_04.tex#Difference of
squares} first 
\begin{eqnarray*}
\left( x^{3}\right) ^{2}-1^{2} &=&0 \\
\left( x^{3}+1\right) \left( x^{3}-1\right)  &=&0
\end{eqnarray*}
So solutions to either $x^{3}+1=0$ or $x^{3}-1=0$ will provide our roots.
First, for $x^{3}+1=0$, use the \hyperref{sum of cubes}{}{}{Al01_04.tex#Sum
of cubes} formula 
\begin{eqnarray*}
x^{3}+1 &=&0 \\
\left( x+1\right) \left( x^{2}-x+1\right)  &=&0
\end{eqnarray*}
So either $x=-1$ or $x^{2}-x+1=0$ will do, and you can solve the second by
the \hyperref{quadratic formula}{}{}{Al02_03.tex#Quadratic Formula} 
\begin{eqnarray*}
x^{2}-x+1 &=&0 \\
x &=&\frac{-\left( -1\right) \pm \sqrt{\left( -1\right) ^{2}-4\left(
1\right) \left( 1\right) }}{2\left( 1\right) } \\
&=&\frac{1\pm \sqrt{5}}{2}
\end{eqnarray*}
Now for the other half of our solutions, look at $x^{3}-1=0$, using the 
\hyperref{difference of cubes}{}{}{Al01_04.tex#Difference of cubes} formula 
\begin{eqnarray*}
x^{3}-1 &=&0 \\
\left( x-1\right) \left( x^{2}+x+1\right)  &=&0
\end{eqnarray*}

So either $x=1$ or $x^{2}+x+1=0$, and you can solve the second by the 
\hyperref{quadratic formula}{}{}{Al02_03.tex#Quadratic Formula} 
\begin{eqnarray*}
x^{2}+x+1 &=&0 \\
x &=&\frac{-1\pm \sqrt{1^{2}-4\left( 1\right) \left( 1\right) }}{2\left(
1\right) } \\
&=&\frac{-1\pm \sqrt{-3}}{2} \\
&=&-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}
\end{eqnarray*}
So the complete list of roots is 
\[
1\text{,}-1\text{,}\frac{1+\sqrt{5}}{2}\text{,}\frac{1-\sqrt{5}}{2}\text{,}-%
\frac{1}{2}+i\frac{\sqrt{3}}{2}\text{,}-\frac{1}{2}-i\frac{\sqrt{3}}{2}
\]
\end{itemize}

\subsubsection{Factoring by Grouping\label{Factoring by grouping}}

\begin{itemize}
\item[Example]  Find all solutions of 
\[
12x^{3}-4x^{2}+3x-1=0
\]

\item[Solution]  You can factor the left-hand side of the equation by
grouping the terms in pairs 
\begin{eqnarray*}
\left( 12x^{3}-4x^{2}\right) +\left( 3x-1\right)  &=&0 \\
\left( 4x^{2}\right) \left( 3x-1\right) +\left( 1\right) \left( 3x-1\right) 
&=&0 \\
\left( 4x^{2}+1\right) \left( 3x-1\right)  &=&0 \\
4x^{2}+1 &=&0\text{ or }3x-1=0 \\
x^{2} &=&-\frac{1}{4}\text{ or }3x=1 \\
x &=&\pm \frac{1}{2}i\text{ or }x=\frac{1}{3}
\end{eqnarray*}
\end{itemize}

\subsubsection{Factoring involving Fractional Exponents\label%
{Factoring involving Fractional Exponents}\hfill \hyperref{\fbox{See also}}{%
}{}{Al01_03.tex#Definition of rational exponents}}

\begin{itemize}
\item[Example]  Find all solutions of 
\[
x^{\frac{5}{6}}+x^{\frac{2}{3}}=2x^{\frac{1}{2}}
\]

\item[Solution]  First, move all terms to the left hand side 
\[
x^{\frac{5}{6}}+x^{\frac{2}{3}}-2x^{\frac{1}{2}}=0
\]
Next, write the exponents using a common denominator ($6$) 
\[
x^{\frac{5}{6}}+x^{\frac{4}{6}}-2x^{\frac{3}{6}}=0
\]
Now factor out the lowest power of $x$ (namely $x^{\frac{3}{6}}$) 
\[
x^{\frac{3}{6}}\left( x^{\frac{2}{6}}+x^{\frac{1}{6}}-2\right) =0\qquad 
\CustomNote[\fbox{Caution}]{Note}{\emph{Remember:} $x^{a}\cdot x^{b}=x^{a+b}$%
, so factoring $x^{a}$ out of $x^{a+b}$ leaves a quotient of $x^{b}$. This
seems obvious with natural number exponents, but is easily forgotten when
dealing with other types of exponent.}\text{ }
\]
\end{itemize}

\noindent \hrulefill

\begin{center}
\begin{tabular}{ccc}
{\small \hyperref{\FRAME{itbpF}{13.9375pt}{14.4375pt}{3pt}{}{}{nblogo.wmf}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 13.9375pt;height 14.4375pt;depth
3pt;original-width 24.0625pt;original-height 24.8125pt;cropleft "0";croptop
"0.9959";cropright "0.9937";cropbottom "0";filename
'Nblogo.wmf';file-properties "XNPEU";}}Computing Techniques}{}{}{%
../../Help/DMindex.tex}} & {\small \hyperref{\FRAME{itbpF}{13.9375pt}{%
14.4375pt}{3pt}{}{}{nblogo.wmf}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
13.9375pt;height 14.4375pt;depth 3pt;original-width
24.0625pt;original-height 24.8125pt;cropleft "0";croptop "0.9959";cropright
"0.9937";cropbottom "0";filename 'Nblogo.wmf';file-properties "XNPEU";}}%
General Information}{}{}{../../Help/Index.tex}} & {\small \hyperref{\FRAME{%
itbpF}{13.9375pt}{14.4375pt}{3pt}{}{}{nblogo.wmf}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"PICT";valid_file "F";width 13.9375pt;height 14.4375pt;depth
3pt;original-width 24.0625pt;original-height 24.8125pt;cropleft "0";croptop
"0.9959";cropright "0.9937";cropbottom "0";filename
'Nblogo.wmf';file-properties "XNPEU";}}Reference Library}{}{}{%
../ReferenceIndex.tex}}
\end{tabular}
\end{center}

\noindent \hrulefill

\end{document}
