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%TCIDATA{<META NAME="Title" CONTENT="Exercises">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:13:37 1997}
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\input{tcilatex}
\begin{document}


\section{Exercises\label{Exercises}}

\begin{enumerate}
\item[1.]  \label{q1}Given that when $x^{2}-3x+5k$ is divided by $x+4$ the
remainder is $9$, find the value of $k$ using \textsf{Divide} on the \textsf{%
Polynomials} submenu and \textsf{Solve + Exact}.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a1}

\item[2.]  \label{q2}Define functions $f(x)=x^{3}+x\ln x$ and $g(x)=x+e^{x}$%
. Evaluate $f(g(x))$, $g(f(x))$, $f(x)g(x)$, and $f(x)+g(x)$.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a2}

\item[3.]  \label{q3}Find the equation of the line passing through the two
points $\left( x_{1},y_{1}\right) $, $\left( x_{2},y_{2}\right) $.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a3}

\item[4.]  \label{q4}Find the equation of the line passing through the two
points $\left( 2,5\right) $, $\left( 3,-7\right) $.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a4}

\item[5.]  \label{q5}Find the equation of the line passing through the two
points $\left( 1,2\right) $, $\left( 2,4\right) $.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a5}

\item[6.]  \label{q6}Find the slope of the line determined by the equation $%
sx+ty=c$.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}%
}{}{}{a6}

\item[7.]  \label{q7}Find the center and semi-axes of the ellipse $%
16\,x^{2}+4y^{2}+96x-8y+84=0$.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{%
0.2171in}{0.2006in}{0in}}{}{}{a7}

\item[8.]  \label{q8}Factor the difference of powers $x^{n}-y^{n}$ for
several values of $n$, and deduce a general formula.\hfill \hyperref{%
\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a8}

\item[9.]  \label{q9}Applying \textsf{Factor} to $x^{2}+\left( \sqrt{5}%
-3\right) x-3\sqrt{5}$ gives the factorization 
\[
x^{2}+\left( \sqrt{5}-3\right) x-3\sqrt{5}=\,\left( x+\sqrt{5}\right) \left(
x-3\right) 
\]
showing that the system can factor some polynomials with irrational roots.
However, applying \textsf{Factor} to $\,x^{2}-3$ and $x^{3}+3x^{2}-5x+1$
does not do anything. Find a way to get \textsl{Scientific Notebook} to
factor these polynomials.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in%
}{0.2006in}{0in}}{}{}{a9}

\item[10.]  \label{q10}If principal $P$ is invested at an interest rate $r$
compounded annually, in $t$ years it grows to an amount $A$ given by $%
A=P\left( 1+r\right) ^{t}$. Find the number of years required for the amount
to double at interest rates of $3\%$, $5\%$, $8\%$, and $10\%$.

\begin{itemize}
\item  Analytic solution: Solving the appropriate equation for $t$, define $r
$ to be each of the different interest rates, and evaluate the expression
you obtained for $t$.

\item  Experimental solution: Treat $A$ as a function of $r$ and $t$, define
different values for $r$ and $t,$ and make an appropriate table of values
for each interest rate $r$.
\end{itemize}

\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a10}

\item[11.]  \label{q11}Find all real and complex solutions to the system of
equations 
\begin{eqnarray*}
2x^{2}-y &=&1 \\
x+3y^{3} &=&4
\end{eqnarray*}

\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a11}

\item[12.]  \label{q12}\label{Conic sections}\label{Discriminant}The curves
determined by quadratic equations 
\index{Quadratic equation@Quadratic equation}\label{Quadratic equation}of
the form $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$ are often referred to as \emph{conic
sections}, as each can be described as the intersection of a plane with a
cone. The \emph{discriminant} 
\index{Discriminant@Discriminant}of the expression $%
Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F $ is the number 
\[
\bigtriangleup \,=B^{2}-4AC 
\]
The type of curve determined by the quadratic equation depends on the \emph{%
sign} of the discriminant. In nondegenerate cases, the quadratic equation
gives 
\[
\fbox{$
\begin{array}{r}
\text{an ellipse if }\bigtriangleup \;<\;0 \\ 
\text{a parabola if }\bigtriangleup \;=\;0 \\ 
\text{a hyperbola if }\bigtriangleup \;>\;0
\end{array}
$} 
\]
where a circle is considered as a special case of an ellipse. The degenerate
cases give a pair of lines, a line, or a point, or they have imaginary
solutions.

A rotation of one or both axes in the plane leads to a change of variables
such as the change 
\begin{eqnarray*}
u &=&ax+by \\
v &=&cx+dy
\end{eqnarray*}
from $\left( x,y\right) $ to $\left( u,v\right) $. Show that the sign of the
discriminant (and thus the basic shape of the curve) is invariant under such
changes for a quadratic equation $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$ if $ad\neq bc$%
.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2171in}{0.2006in}{0in}}{}{}{a12%
}
\end{enumerate}

\strut

\section{Solutions\label{Solutions}}

\begin{enumerate}
\item[1.]  \label{a1}Using \textsf{Divide} on the \textsf{Polynomials}
submenu, 
\[
\frac{x^{2}-3x+5k}{x+4}=\,x-7+\frac{5k+28}{x+4} 
\]
Thus, the remainder is $5k+28$. Applying \textsf{Solve + Exact} to $5k+28=9$
gives the solution $\left\{ k=-\frac{19}{5}\right\} $.\hfill \hyperref{%
\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q1}

\item[2.]  \label{a2}Defining functions $f(x)=x^{3}+x\ln x$ and $%
g(x)=x+e^{x} $ and evaluating gives 
\begin{eqnarray*}
f(g(x)) &=&\left( x+e^{x}\right) ^{3}+\left( x+e^{x}\right) \ln \left(
x+e^{x}\right) \\
g(f(x)) &=&x^{3}+x\ln x+e^{x^{3}+x\ln x} \\
f(x)g(x) &=&\left( x^{3}+x\ln x\right) \left( x+e^{x}\right) \\
f(x)+g(x) &=&x^{3}+x\ln x+x+e^{x}
\end{eqnarray*}

\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q2}

\item[3.]  \label{a3}For any two distinct points $(x_{1},y_{1})$ and $%
(x_{2},y_{2})$ in the plane, there is a unique line $ax+by+c=0$ through
these two points. Substituting these points in the equation for the line
gives the two equations $ax_{1}+by_{1}+c=0$ and $ax_{2}+by_{2}+c=0$. Apply 
\textsf{Exact} from the \textsf{Solve} submenu to this system 
\begin{eqnarray*}
ax_{1}+by_{1}+c &=&0 \\
ax_{2}+by_{2}+c &=&0
\end{eqnarray*}
of linear equations, solving for the variables $a,b$. 
\[
\text{Solution is: }\left\{ b=-c\dfrac{x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}%
,a=c\dfrac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}\right\} 
\]
Consequently, the equation for the line is 
\[
c\left( \frac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}\right) x-c\left( \frac{%
x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}\right) y+c=0 
\]
or, clearing fractions and collecting coefficients by factoring in place, 
\[
\left( y_{1}-y_{2}\right) x-\left( x_{1}-x_{2}\right) y+\left(
x_{1}y_{2}-y_{1}x_{2}\right) =0 
\]

\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q3}

\item[4.]  \label{a4}For the points $\left( 2,5\right) $, $\left(
3,-7\right) $, the system of equations is 
\begin{eqnarray*}
2a+5b+c &=&0 \\
3a-7b+c &=&0
\end{eqnarray*}
Apply \textsf{Exact} from the \textsf{Solve} submenu to get 
\[
\text{Solution is: }\left\{ b=-\frac{1}{29}c,a=-\frac{12}{29}c\right\} 
\]
Consequently, the equation for the line is 
\[
-\frac{12}{29}cx-\left( -\frac{1}{29}\right) cy+c=0 
\]
or, clearing fractions and simplifying, 
\[
-12x+y+29=0 
\]

\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q4}

\item[5.]  \label{a5}Since the point $\left( 0,0\right) $ lies on the line,
you do not get a unique solution to the system of equations for the pair $%
a,b $. Thus, choosing \textsf{Exact} from the \textsf{Solve} submenu and
specifying $a,b$ for the variables gives no response. However, specifying $%
a,c$ for \textsf{Variable(s) to Solve for} gives the solution 
\[
\left\{ a=-2b,c=0\right\} 
\]
Thus, the equation for the line is 
\[
-2bx+by=0 
\]
or, dividing by $b$ and applying \textsf{Simplify}, 
\[
\left( -2bx+by\right) \frac{1}{b}=-2x+y=0 
\]
\emph{Note}: An interesting method for finding the equation of a line
through two specified points using determinants is described in Chapter 5,
Matrix Algebra.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{%
0.243in}{0in}}{}{}{q5}

\item[6.]  \label{a6}The \emph{slope-intercept} form of the equation for a
line is $y=mx+b$, where $m$ is the slope and $b$ the $y$-intercept. If a
line is given as a linear equation in the form 
\[
sx+ty=c 
\]
you can find the slope by solving the equation for $y$. Apply \textsf{Expand}
to the solution 
\[
y=-\frac{sx-c}{t} 
\]
to get 
\[
y=-\frac{1}{t}sx+\frac{1}{t}c 
\]
revealing the slope to be $-\dfrac{s}{t}$.\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q6}

\item[7.]  \label{a7}To find the center and semi-axes of the ellipse $%
16x^{2}+4y^{2}+96x-8y+84=0$,

\begin{itemize}
\item  Subtract $84$ from both sides of the equation to get 
\[
\left( 16\,x^{2}+4y^{2}+96x-8y+84\right) -84=0-84 
\]

\item  Select the left side and while holding down the \textsc{ctrl } key,
apply \textsf{Simplify}; then, do the same to the right side to get 
\[
16x^{2}+4y^{2}+96x-8y=-84 
\]

\item  Drag the terms containing $x$ together, select them and click \FRAME{%
itbpF}{0.2992in}{0.2992in}{0.0692in}{}{}{parens.wmf}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"PICT";valid_file "F";width 0.2992in;height 0.2992in;depth
0.0692in;original-width 0pt;original-height 0pt;cropleft "0";croptop
"1.0204";cropright "1.0676";cropbottom "0";filename
'Parens.wmf';file-properties "XNPEU";}}; then, do the same to the terms
containing $y$, to get 
\[
\left( 16x^{2}+96x\right) +\left( 4y^{2}-8y\right) =-84 
\]

\item  Factor out the leading coefficients by dragging them outside the
parentheses and dividing other coefficients by their value, to get 
\[
16\left( x^{2}+\frac{96}{16}x\right) +4\left( y^{2}-\frac{8}{4}y\right) =-84 
\]

\item  Add the product of the coefficient of $x^{2}$ with the square of
one-half the coefficient of $x$ to both sides; then, do the same for $y$, to
get 
\[
\begin{array}{c}
16\left( x^{2}+\frac{96}{16}x+\left( \frac{1}{2}\frac{96}{16}\right)
^{2}\right) +4\left( y^{2}-\frac{8}{4}y+\left( \frac{1}{2}\frac{8}{4}\right)
^{2}\right) \vspace{6pt} \\ 
=-84+16\left( \frac{1}{2}\frac{96}{16}\right) ^{2}+4\left( \frac{1}{2}\frac{8%
}{4}\right) ^{2}\vspace{6pt}
\end{array}
\]

\item  Select the term $16\left( x^{2}+\frac{96}{16}x+\left( \frac{1}{2}%
\times \frac{96}{16}\right) ^{2}\right) $ and while holding down the \textsc{%
ctrl } key, apply \textsf{Factor}; then, do the same for the term with $y$,
to get 
\[
16\left( x+3\right) ^{2}+4\left( y-1\right) ^{2}=-84+16\left( \frac{1}{2}%
\frac{96}{16}\right) ^{2}+4\left( \frac{1}{2}\frac{8}{4}\right) 
\]

\item  Select the right side and while holding down the \textsc{ctrl } key,
apply \textsf{Simplify}, to get 
\[
16\left( x+3\right) ^{2}+4\left( y-1\right) ^{2}=64 
\]

\item  Divide each term by the right-hand side, to get 
\[
\frac{16\left( x+3\right) ^{2}}{64}+\frac{4\left( y-1\right) ^{2}}{64}=\frac{%
64}{64} 
\]

\item  Select each term and while holding down the \textsc{ctrl } key, apply 
\textsf{Factor}, to get 
\[
\frac{1}{4}\left( x+3\right) ^{2}+\frac{1}{16}\left( y-1\right) ^{2}=1 
\]
\end{itemize}

You can read the answer from this form of the equation: The center of the
ellipse is $(-3,1)$, and the semi-axes are $\sqrt{4}=\,2$ and $\sqrt{16}=\,4$%
.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q7}

\item[8.]  \label{a8}Apply \textsf{Factor} to several differences.

\begin{description}
\item  $x^{2}-y^{2}=\,\left( x-y\right) \left( x+y\right) $

\item  $x^{3}-y^{3}=\,\left( x-y\right) \left( x^{2}+xy+y^{2}\right) $

\item  $x^{4}-y^{4}=\,\left( x-y\right) \left( x+y\right) \left(
x^{2}+y^{2}\right) $

\item  $x^{5}-y^{5}=\,\left( x-y\right) \left(
x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}\right) $

\item  $x^{6}-y^{6}=\,\left( x-y\right) \left( x+y\right) \left(
x^{2}+xy+y^{2}\right) \left( x^{2}-xy+y^{2}\right) $

\item  $x^{7}-y^{7}=\,\left( x-y\right) \left(
x^{6}+x^{5}y+x^{4}y^{2}+x^{3}y^{3}+x^{2}y^{4}+xy^{5}+y^{6}\right) $
\end{description}

After looking at only these few examples, you might find it reasonable to
conjecture that, for $n$ odd, 
\[
x^{n}-y^{n}=\,\left( x-y\right) \sum_{k=0}^{n-1}x^{n-k-1}y^{k} 
\]
We leave the general conjecture for you. Experiment.\hfill \hyperref{%
\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q8}

\item[9.]  \label{a9}Using the clue from the example that the system will
factor over roots that appear as coefficients, apply \textsf{Factor} to the
product $\sqrt{3}\left( x^{2}-3\right) $ to get $\sqrt{3}\left(
x^{2}-3\right) =\,\sqrt{3}\left( x-\sqrt{3}\right) \left( x+\sqrt{3}\right) $%
. Now you can divide out the extraneous factor to get 
\[
x^{2}-3=\,\left( x-\sqrt{3}\right) \left( x+\sqrt{3}\right) 
\]

For the polynomial $x^{3}+3x^{2}-5x+1$, apply \textsf{Roots} from the 
\textsf{Polynomials} submenu to find the roots: $\left\{ 1,-2+\sqrt{5},-2-%
\sqrt{5}\right\} $. You can multiply by $\sqrt{5}$ to factor this
polynomial: $\sqrt{5}\left( x^{3}+3x^{2}-5x+1\right) \,=\,\sqrt{5}\left(
x-1\right) \left( x+2+\sqrt{5}\right) \left( x+2-\sqrt{5}\right) $. Then,
dividing out the extraneous factor of $\sqrt{5},$ you have 
\[
x^{3}+3x^{2}-5x+1\,=\,\left( x-1\right) \left( x+2+\sqrt{5}\right) \left(
x+2-\sqrt{5}\right) 
\]

\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q9}

\item[10.]  \label{a10}\textbf{Analytic solution:} The question asks when $%
A=2P$. Solving the equation $2P=P\left( 1+r\right) ^{t}$ for $t$ gives the
solution 
\[
t=\frac{\ln 2}{\ln \left( 1+r\right) } 
\]
\textsf{Evaluate Numerically} gives the number of years for each of the four
different interest rates as follows. 
\[
\fbox{$
\begin{array}{c}
\text{For 3\%, \ }\dfrac{\ln 2}{\ln \left( 1+.03\right) }=23.4\text{ years;
For 5\%, \ }\dfrac{\ln 2}{\ln \left( 1+.05\right) }=14.2\text{ years\vspace{%
12pt}} \\ 
\text{For 8\%,\ \ }\dfrac{\ln 2}{\ln \left( 1+.08\right) }=9.01\text{ years;
For 10\%, }\dfrac{\ln 2}{\ln \left( 1+.10\right) }=7.27\text{ years}
\end{array}
$} 
\]
$\,\,$

\textbf{Experimental solution:} For the interest rate $3\%$, you might start
with the following data: 
\[
\fbox{$
\begin{array}{cc}
10\text{ years:} & A=P\left( 1+.03\right) ^{10}=1.34P \\ 
15\text{ years:} & A=P\left( 1+.03\right) ^{15}=1.56P \\ 
20\text{ years:} & A=P\left( 1+.03\right) ^{20}=\,1.81P \\ 
25\text{ years:} & A=P\left( 1+.03\right) ^{25}=\,2.09P
\end{array}
$} 
\]
Then, look at the years between 20 and 25:$\,$ 
\[
\fbox{$
\begin{array}{cc}
21\text{ years:} & A=P\left( 1+.03\right) ^{21}=1.86P \\ 
22\text{ years:} & A=P\left( 1+.03\right) ^{22}=\,1.92P \\ 
23\text{ years:} & A=P\left( 1+.03\right) ^{23}=\,1.97P \\ 
24\text{ years:} & A=P\left( 1+.03\right) ^{24}=\,2.03P
\end{array}
$} 
\]
These results indicate that it takes between 23 and 24 years to double an
investment at 3\% annual interest. Take similar steps for the other interest
rates.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in%
}}{}{}{q10}

\item[11.]  \label{a11}With the insertion point in the array, from the 
\textsf{Solve} submenu, choose \textsf{Exact}. You receive the response 
\[
\begin{array}{c}
\text{Solution is : }\left\{ y=1,x=1\right\} ,\,\left\{ y=\rho ,x=-3\rho
^{3}+4\right\} \\ 
\text{where }\rho \text{ is a root of }\left(
18Z^{5}+18Z^{4}+18Z^{3}-30Z^{2}-30Z-31\right)
\end{array}
\]

Leave your insertion point in the polynomial 
\[
18Z^{5}+18Z^{4}+18Z^{3}-30Z^{2}-30Z-31 
\]
and from the \textsf{Polynomials} submenu, choose \textsf{Roots}. You get
the following solution. 
\[
\text{roots}:\fbox{$
\begin{array}{c}
-.60663-.98268i \\ 
-.60663+.98268i \\ 
-.48801-.92069i \\ 
-.48801+.92069i \\ 
1.1893
\end{array}
$} 
\]
Define the function $x(t)=-3t^{3}+4$ with \textsf{New Definition} from the 
\textsf{Define} submenu. Select the vector of roots, and click the
parentheses icon. Type an $x$ at the left of the vector, leave the insertion
point in the expression, and apply \textsf{Evaluate} to get the following: 
\[
x\left( 
\begin{array}{c}
-.60663-.98268i \\ 
-.60663+.98268i \\ 
-.48801-.92069i \\ 
-.48801+.92069i \\ 
1.1893
\end{array}
\right) =\,\left( 
\begin{array}{c}
-3\left( -.60663-.98268i\right) ^{3}+4 \\ 
-3\left( -.60663+.98268i\right) ^{3}+4 \\ 
-3\left( -.48801-.92069i\right) ^{3}+4 \\ 
-3\left( -.48801+.92069i\right) ^{3}+4 \\ 
-1.0466
\end{array}
\right) 
\]
Leave the insertion point in the matrix on the right and click \textsf{Expand%
}. 
\[
\left( 
\begin{array}{c}
-3\left( -.60663-.98268i\right) ^{3}+4 \\ 
-3\left( -.60663+.98268i\right) ^{3}+4 \\ 
-3\left( -.48801-.92069i\right) ^{3}+4 \\ 
-3\left( -.48801+.92069i\right) ^{3}+4 \\ 
-1.0466
\end{array}
\right) =\,\left( 
\begin{array}{c}
-.60247+.40783i \\ 
-.60247-.40783i \\ 
.62562-.36793i \\ 
.62562+.36793i \\ 
-1.0466
\end{array}
\right) 
\]
To display this result, you can concatenate the two vectors: Place them side
by side and from the \textsf{Matrices} submenu, choose \textsf{Concatenate}.
Then, select the (two-column) matrix, choose \textsf{Edit + Insert Row(s)}
to add a new first row, and label the columns with $x$ and $y$ as follows: 
\[
\fbox{$
\begin{array}{cc}
\mathbf{x=-3y}^{3}\mathbf{+4} & \mathbf{y} \\ 
-.60247+.40783i & -.60663-.98268i \\ 
-.60247-.40783i & -.60663+.98268i \\ 
.62562-.36793i & -.48801-.92069i \\ 
.62562+.36793i & -.48801+.92069i \\ 
-1.0466 & 1.1893
\end{array}
$} 
\]

\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{%
}{q11}

\item[12.]  \label{a12}First, solve the system of equations for $x$ and $y$
with \textsf{Solve + Exact} to get the solution 
\[
x=-\dfrac{bv-du}{ad-bc}\text{\quad \quad }y=\dfrac{av-cu}{ad-bc} 
\]
(Note that the condition $ad\neq bc$ is necessary for this system of
equations to have a solution.) Now find the quadratic equation in $u$ and $v$%
. To replace $x$ and $y$ with their equivalent expressions in terms of $u$
and $v$, edit the polynomial; or define the variables $x$ and $y$ and apply 
\textsf{Evaluate}; or select the polynomial and from the \textsf{Edit} menu,
choose \textsf{Replace} and enter the appropriate expressions in
mathematics. These replacements get the following polynomial: 
\[
\begin{array}{c}
A\left( -\dfrac{vb-du}{-cb+ad}\right) ^{2}+B\left( -\dfrac{vb-du}{-cb+ad}%
\right) \left( \dfrac{-cu+av}{-cb+ad}\right) +C\left( \dfrac{-cu+av}{-cb+ad}%
\right) ^{2}\vspace{6pt} \\ 
+D\left( -\dfrac{vb-du}{-cb+ad}\right) +E\left( \dfrac{-cu+av}{-cb+ad}%
\right) +F=0
\end{array}
\]
To simplify this equation and get it into standard quadratic form in terms
of the variables $u$ and $v$, first factor the left side of the equation to
find a common denominator. (This procedure gives a fraction too large to fit
well on the page 
\[
-\frac{-Av^{2}b^{2}+2Avbdu-Ad^{2}u^{2}-\cdots -Fc^{2}b^{2}+2Fcbad-Fa^{2}d^{2}%
}{\left( ad-cb\right) ^{2}} 
\]
but you can still work with it on your screen.) Multiply by the denominator $%
\left( ad-cb\right) ^{2}$ to get the following quadratic equation in powers
of $u$ and $v$: 
\[
\begin{tabular}{l}
$Av^{2}b^{2}-2Avbdu+Ad^{2}u^{2}+Bvbcu-Bv^{2}ba-Bdu^{2}c$ \\ 
$\qquad +\,Bduav+Cc^{2}u^{2}-2Ccuav+Ca^{2}v^{2}+Dvb^{2}c$ \\ 
$\qquad -Dvbad-\,Dducb+Dd^{2}ua+Ec^{2}ub-Ecuad$ \\ 
$\qquad -Eavcb+Ea^{2}vd+\,Fc^{2}b^{2}-2Fcbad+Fa^{2}d^{2}=0$%
\end{tabular}
\]
$\,$Now, to identify the sign of the discriminant for this quadratic
equation, you need to collect coefficients of $u$ and $v$. To collect them,
select and then click and drag terms containing a factor of $u^{2}$ to the
beginning of the expression. Then, select all these terms, and while holding
down the \textsc{ctrl } key, choose \textsf{Factor}. Then, follow this same
procedure for $uv$, $v^{2}$, $u$, and $v$ in turn. You can get this equation
into the following form with these steps: 
\[
\begin{tabular}{l}
$\left( d^{2}A-Bcd+c^{2}C\right) u^{2}+\left( -2bdA+aBd+bBc-2acC\right) uv$
\\ 
$\qquad +\left( -abB+Ab^{2}+a^{2}C\right) v^{2}+\left( ad-cb\right) \left(
Dd-cE\right) u$ \\ 
$\qquad +\left( ad-cb\right) \left( Ea-Db\right) v+\,F\left( -cb+ad\right)
^{2}=0$%
\end{tabular}
\]
To find the discriminant, replace $\widetilde{A},\widetilde{B},\widetilde{C}$
in the expression $\widetilde{\bigtriangleup }=\widetilde{B}^{2}-4\widetilde{%
A}\widetilde{C}$ with these coefficients: 
\begin{eqnarray*}
\widetilde{A} &=&d^{2}A-Bcd+c^{2}C \\
\widetilde{B} &=&-2bdA+aBd+bBc-2acC \\
\widetilde{C} &=&-abB+Ab^{2}+a^{2}C
\end{eqnarray*}
By clicking and dragging, move terms having a factor of $B^{2}$ adjacent to
one another. Repeat for terms having a factor of $AC$. Then, select all
terms having factors of $B^{2}$ and, holding down the \textsc{ctrl } key,
factor in place. Repeat for all terms having factors of $AC$. Then, factor
the result. These successive steps are demonstrated in the following
display: 
\begin{eqnarray*}
\widetilde{\bigtriangleup } &=&\widetilde{B}^{2}-4\widetilde{A}\widetilde{C}
\\
\qquad &=&\left( -2bdA+aBd+bBc-2acC\right) ^{2} \\
&&\qquad \qquad -\,4\left( d^{2}A-Bcd+c^{2}C\right) \left(
-abB+Ab^{2}+a^{2}C\right) \\
\qquad
&=&%
\,8bdAacC+a^{2}B^{2}d^{2}-2aB^{2}dbc+b^{2}B^{2}c^{2}-4d^{2}Aa^{2}C-4c^{2}CAb^{2}
\\
\qquad
&=&a^{2}B^{2}d^{2}-2aB^{2}dbc+b^{2}B^{2}c^{2}+8bdAacC-4d^{2}Aa^{2}C-4c^{2}CAb^{2}
\\
\qquad &=&B^{2}\left( ad-cb\right) ^{2}-4AC\left( ad-cb\right) ^{2} \\
\qquad &=&\left( B^{2}-4AC\right) \left( ad-cb\right) ^{2} \\
\qquad &=&\bigtriangleup \left( ad-cb\right) ^{2}
\end{eqnarray*}
It is easy to see from the equation 
\[
\widetilde{\bigtriangleup }=\bigtriangleup \left( ad-cb\right) ^{2} 
\]
that $\bigtriangleup $ and $\widetilde{\bigtriangleup }$ have the same sign
when $ad\neq bc$.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{%
0.243in}{0in}}{}{}{q12}
\end{enumerate}

\strut

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Polynomials and Rational Expressions}{}{}{%
DM3-1.tex#Polynomials}

\item  \hyperref{Defining Variables and Functions}{}{}{DM3-2.tex#Define}

\item  \hyperref{Solving Polynomial Equations}{}{}{DM3-3.tex#Exact solutions}

\item  \hyperref{Substitution}{}{}{DM3-4.tex#Substitution}

\item  \hyperref{Exponents and Logarithms}{}{}{DM3-5.tex#Exponential function%
}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{38.875pt}{24.5pt}{6pt}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid-file "F";width 38.875pt;height 24.5pt;depth
6pt;original-width 36.125pt;original-height 22.5625pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'index.wmf';file-properties
"XNPEU";}}}{}{}{DMIndex.tex#Math Index}\quad {\small Index entries: 
\hyperref{Conic sections}{}{}{DMIndex.tex#Conic sections}, \hyperref{%
Discriminant}{}{}{DMIndex.tex#Discriminant}, \hyperref{Quadratic equations}{%
}{}{DMIndex.tex#Quadratic equations}}

\end{document}
