\magnification=\magstep1
\vsize=9truein
\hsize=6.5truein
%\parindent=0.8truein
\def\C{{\bf C}}
\def\R{{\bf R}}
\def\N{{\bf N}}
\def\span{{\rm span}}
\font\sc=cmcsc10
\font\smalltt=cmtt10 scaled840
\font\smseven=cmr7 scaled840

%verbatim environment
\def\uncatcodespecials{\def\do##1{\catcode`##1=12 }\dospecials}
\newcount\lineno
%\def\listing#1{\par\begingroup\setupverbatim #1\endgroup}
\def\setupverbatimc{\parindent=1truecm\tt
	\smalltt\baselineskip=10pt plus 1pt minus 1pt
	\def\par{\leavevmode\endgraf}\catcode`\`=\active\obeylines
	\uncatcodespecials\obeyspaces
	\everypar{\global\advance\lineno by 1\llap{\smseven\the\lineno\ \ }}}
\def\setupverbatimb{\parindent=1truecm\global\lineno=0%
	\smalltt\baselineskip=10pt plus 1pt minus 1pt
	\def\par{\leavevmode\endgraf}\catcode`\`=\active\obeylines
	\uncatcodespecials\obeyspaces
	\everypar{\advance\lineno by 1{}}}
\def\setupverbatim{\global\lineno=0\setupverbatimc}

{\obeyspaces\global\let =\ }

\def\beginverbatim{\par\begingroup\setupverbatim\doverbatim}
{\catcode`\|=0\catcode`\\=12\catcode`|`=|active|gdef`{|relax|lq}%
|obeylines|gdef|doverbatim^^M#1\endverbatim{#1|endgroup}}

\def\beginverbatimb{\par\begingroup\setupverbatimb\doverbatim}
\def\listing#1{\medskip
	\begingroup\setupverbatim\input#1\endgroup\medskip
	\noindent\ignorespaces}
\def\listout#1{\medskip
	\begingroup\setupverbatimb\input#1\endgroup\medskip
	\noindent\ignorespaces}
\def\macinput#1{{\catcode`\$=10\catcode`@=3 \input #1 }}

\newcount\qnum\qnum=0
\def\qn{\advance\qnum by1\medskip\par\noindent\item{\the\qnum.}}

\input epsf

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\qn [Kincaid and Cheney Problem 3.2\#4] \
Steffensen's method is given by the iteration
$$
    x_{n+1}=x_n-{f(x_n)\over g(x_n)}
\qquad\hbox{where}\qquad
    g(x)={f\big(x+f(x)\big)-f(x)\over f(x)}.
$$
Show this method is quadratically convergent under suitable hypothesis.
\medskip\noindent
Let $r$ be a solution such that $f(r)=0$ and $f'(r)\ne 0$.
Moreover assume $|f''|$ is bounded in a neighborhood of $r$.
We shall show that Steffensen's method is quadratically convergent
provided that $x_0$ is sufficiently close to $r$.

Consider the function
$$
	\kappa (x,\xi,\eta)=
	    {f''(\xi)\big(f'(x)
        -{\textstyle {1\over 2}} f''(\eta)(x-r)\big)+f''(\eta)
    \over 2f'(x)+f''(\xi)f(x)}.
$$
By assumption the limit supremum of $\big|\kappa(x,\xi,\eta)\big|$ 
exists as $x\to r$, $\xi\to r$ and $\eta\to r$.  Let $C$ denote
this limit and choose $\delta_1>0$ so small that
$\big|\kappa(x,\xi,\eta)\big|\le 2C$ for every $x$, $\xi$ and $\eta$
such that
$|x-r|<\delta_1$, $|\xi-r|<\delta_1$ and $|\eta-r|<\delta_1$.
Choose $\delta_2>0$ so small that 
$|x-r|<\delta_2$ implies
$2\big|f(x)\big|\le \delta_1$.
Now let 
$$\delta=\min\Big\{{\delta_1\over  2},\delta_2,{1\over 2C}\Big\}.$$
Claim that if $x_0\in (r-\delta,r+\delta)$ then $x_n$ defined
according to Steffensen's method satisfies $x_n\in (r-\delta,r+\delta)$
for all $n\in\N$.


Let $x_0\in (r-\delta,r+\delta)$ and define $x_n$ according
to Steffensen's method.  
By Taylor's theorem 
there exists $\xi_n$ between $x_n$ and $x_n+f(x_n)$ such that
$$
	f\big(x_n+f(x_n)\big)=f(x_n)+f'(x_n)f(x_n)
		+{\textstyle {1\over 2}}f''(\xi_n)f(x_n)^2.
$$
Therefore
$$
	g(x_n)=f'(x_n)+{\textstyle {1\over 2}}f''(\xi_n)f(x_n).$$
It follows that
$$
	e_{n+1}=x_{n+1}-r=e_n-{f(x_n)\over f'(x_n)+{1\over 2}f''(\xi_n)f(x_n)}
$$
where $e_n=x_n-r$.

Using Taylor's series again yields
$$
	0=f(r)=f(x_n-e_n)
		=f(x_n)-f'(x_n)e_n+{\textstyle {1\over 2}} f''(\eta_n)e_n^2
$$
for some choice of $\eta_n$ between $r$ and $x_n$.  Therefore
$$\eqalign{
	e_{n+1}
%&=e_n-{f'(x_n)e_n+{\textstyle {1\over 2}} f''(\eta_n)e_n^2
%	\over f'(x_n)+{1\over 2}f''(\xi_n)f(x_n)}
&=e_n-e_n{f'(x_n)+{\textstyle {1\over 2}} f''(\eta_n)e_n
	\over f'(x_n)+{1\over 2}f''(\xi_n)f(x_n)}\cr
&=e_n-e_n{f'(x_n)-{1\over 2}f''(\xi_n)f(x_n)
	+{1\over 2}f''(\xi_n)f(x_n)+{\textstyle {1\over 2}} f''(\eta_n)e_n
	\over f'(x_n)+{1\over 2}f''(\xi_n)f(x_n)}\cr
&=-e_n{
	f''(\xi_n)f(x_n)+f''(\eta_n)e_n
	\over 2f'(x_n)+f''(\xi_n)f(x_n)}\cr
&=-e_n{
	f''(\xi_n)\big(f'(x_n)e_n
		-{\textstyle {1\over 2}} f''(\eta_n)e_n^2\big)+f''(\eta_n)e_n
	\over 2f'(x_n)+f''(\xi_n)f(x_n)}\cr
&=-e_n^2
	{f''(\xi_n)\big(f'(x_n)
		-{\textstyle {1\over 2}} f''(\eta_n)e_n\big)+f''(\eta_n)
	\over 2f'(x_n)+f''(\xi_n)f(x_n)}
	=-e_n^2\kappa(x_n,\xi_n,\eta_n).
}$$

Claim that $x_n\in (r-\delta,r+\delta)$ for every $n\in\N$.
For induction suppose $x_n\in (r-\delta,r+\delta)$.  Since
$\xi_n$ is between $x_n$ and $x_n+f(x_n)$ then
$$|\xi_n-r|\le |\xi_n-x_n|+|x_n-r|
		\le |f(x_n)|+|x_n-r|<\delta_2+\delta\le\delta_1.$$
Since $\eta_n$ is between $x$ and $x_n$ then 
$$|\eta_n-r|\le\delta<\delta_1.$$
Thus $x_n$, $\xi_n$ and $\eta_n$ satisfy
$|x_n-r|<\delta_1$, $|\xi_n-r|<\delta_1$ and $|\eta_n-r|<\delta_1$
so consequently $\big|\kappa(x_n,\xi_n,\eta_n)\big|<2C$.  
Therefore 
	$$|x_{n+1}-r|=|e_{n+1}|\le 2Ce_n^2=2C|x_n-r|^2\le 2C\delta^2
		<\delta.$$
This shows $x_{n+1}\in (r-\delta,r+\delta)$ and completes the
induction.  Moreover, we have shown that $x_n$ is quadratically 
convergent.
\vfill\eject

\qn [Kincaid and Cheney Problem 3.2\#6] \ To compute
reciprocals without division, we can solve $x=1/R$ by finding a
zero of the function $f(x)=x^{-1}-R$.  Write a short algorithm
to find $1/R$ by Newton's method applied to $f$.  Do not use
division or exponentiation in your algorithm.
For positive $R$, what starting points are suitable?
\medskip\noindent
Differentiating yields
$$f'(x)=-x^{-2}.$$
Therefore
$$
    \phi(x)=x-{f(x)\over f'(x)}
        =x+{x^{-1}-R\over x^{-2}}
        =x+x-R x^2=x(2-Rx)
$$
leads to the iteration formula
$x_{n+1}=x_{n}(2-R x_n)$.  To determine what starting points
$x_0$ are suitable we look for an interval centered at $1/R$
over which $|\phi'|$ is strictly less than 1.
$$
    \big|\phi'(x)\big|=|2-2Rx|<1
\qquad\hbox{implies}\qquad
    \Big|{1\over R}-x\Big|<{1\over 2R}.
$$
Therefore any starting point
$$
    x_0\in \Big({1\over 2R},{3\over 2R}\Big)
$$
is suitable.  In fact, this interval can be extended to
$$x_0\in\Big(0,{3\over 2R}\Big)$$
because if $x\in(0,1/2R]$ then
$$
    \phi(x)-x=x(2-Rx)-x=x(1-Rx)
$$
implies
$$
    {x\over 2}\le\phi(x)-x<{1\over 2R}.
$$
Consequently, if $x_0\in(0,1/2R]$, then
$x_n$ form an increasing sequence of points that
are eventually contained in the interval where $|\phi'|<1$.
Similarly
$$x_0\in\Big[{3\over 2R},{2\over R}\Big)\qquad\hbox{implies}\qquad
x_1\in\Big(0,{3\over 4R}\Big].$$
It follows that the method iteration will converge for
any 
$$x_0\in\Big(0,{2\over R}\Big).$$

We now procede to find a choice for $x_0$
that does not involve performing any divisions.
Recall that $R$
is stored on the computer using a binary expansion of the form
$$
    R=\sum_{i=1}^{m} b_i 2^{n-i}
        =\sigma 2^n
\qquad\hbox{where}\qquad
    \sigma\in[1/2,1).
$$
Therefore taking
$$
    x_0=\sigma 2^{-n}<{1\over \sigma 2^{n}} = {1\over R}
$$
provides a suitable initial condition by simply changing the
sign of the exponent in the floating point representation
while doing no division or exponentiation to the mantissa.
The following C program implements this approach.
\listing{newtdiv.c}
The output is
\listout{newtdiv.out}
\vfill\eject

\qn [Kincaid and Cheney Problem 3.2\#17] \ Which
of the following sequences converge quadratically.
\smallskip
\centerline{a. $1/n^2$,\quad
b. $1/2^{2^n}$,\quad
c. $1/\sqrt n$,\quad
d. $1/e^n$,\quad
e. $1/n^n$}
\medskip
\noindent
A sequence $x_n$ converging to a limit $L$, converges quadratically
if there is a constant $C$ and an integer $N$ such that
$|x_{n+1}-L|\le C |x_n-L|^2$ for every $n\ge N$.  Since $L=0$ for
each of the sequences in question and each sequence is moreover
positive, then this inequality may be rewritten as
$$
{x_{n+1}\over x_n^2}\le C\qquad\hbox{for every} \qquad n\ge N.
$$
This inequality will hold for some $C$ and $N$ large enough if
and only if
$$
\limsup_{n\to\infty} {x_{n+1}\over x_n^2}
$$
is bounded.  We proceed as follows:
\medskip
$$
\limsup_{n\to\infty} {1/(n+1)^2\over (1/n^2)^2}
	=\lim_{n\to\infty} {n^4\over n^2+2n+1}=\infty
$$
$$
\limsup_{n\to\infty} {1/2^{2^{n+1}}\over (1/2^{2^n})^2}
	=\lim_{n\to\infty} {4^{2^n}\over 4^{2^n}}=1
$$
$$
\limsup_{n\to\infty} {1/\sqrt{n+1}\over (1/\sqrt{n})^2}
	=\lim_{n\to\infty} {n\over\sqrt{n+1}}=\infty
$$
$$
\limsup_{n\to\infty} {1/e^{n+1}\over (1/e^n)^2}
	=\lim_{n\to\infty} {e^{2n}\over e^{n+1}}=
	=\lim_{n\to\infty} e^{n-1}=\infty
$$
$$
\limsup_{n\to\infty} {1/(n+1)^{n+1}\over (1/n^n)^2}
	=\lim_{n\to\infty} {n^{2n}\over (n+1)^{n+1}}\ge
	\lim_{n\to\infty} {n^{2n}\over (2n)^{n+1}}=
	\lim_{n\to\infty} {1\over 4}\Big({n\over 2}\Big)^n=\infty.
$$
Therefore only $1/2^{2^n}$ converges quadratically
while the others don't.


\vfill\eject

\qn [Kincaid and Cheney Computer Problem 3.2\#1] \ Write a computer program
to solve the equation $x=\tan x$ by Newton's method.
Find the roots nearest 4.5 and 7.7.
\medskip\noindent
The program is
\listing{newtontan.c}
and the output is
\listout{newtontan.out}
\vfill\eject

\qn [Kincaid and Cheney Computer Problem 3.2\#8] \ Program the Newton
algorithm in complex arithmetic and test it on these functions
with the given starting points.
\itemitem{a.} $f_1(z)=z^2+1,\qquad z_0=3+i$
\itemitem{b.} $f_2(z)=z+\sin z-3,\qquad z_0=2-i$
\itemitem{c.} $f_3(z)=z^4+z^2+2+3i,\qquad z_0=1$
\medskip\noindent
We shall use the Maple code generation library to 
create the functions $\phi(z)=z-f(z)/f'(z)$ needed for Newton's
iteration.  The Maple script 
\listing{makephi.mpl}
takes an expression $F$ and an integer $N$ and 
creates the file{\tt\ phiN.i }containing C code to compute 
$\phi$ in terms of $z$.
The script can be used to generate code for the functions 
$f_n$ given in the problem with the commands
\medskip
\beginverbatimb
    $ maple -q -DN=1 -DF="z^2+1" makephi.mpl
    $ maple -q -DN=2 -DF="z+sin(z)-3" makephi.mpl
    $ maple -q -DN=3 -DF="z^4+z^2+2+3*I" makephi.mpl
\endverbatim
\medskip\noindent
The resulting files are
\listing{phi1.i}
\listing{phi2.i}
\listing{phi3.i}
Notice that{\tt\ phi3.i }contains an optimization where 
$z^2$ is stored in a temporary variable.
These files are included in our main program{\tt\ newtcomp.c }which 
is given by
\listing{newtcomp.c}
The output is
\listout{newtcomp.out}
\vfill\eject

\qn [Kincaid and Cheney Computer Problem 3.2\#9] \ Use
Steffensen's method to find the first 10 positive roots 
of the equation $x=\tan x$.
\medskip\noindent
First graph the equation $y=\tan x$ and $y=x$ to identify
on what intervals the roots will be.  The gnuplot script
\listing{xtanx.plt}
produces the output

\centerline{\input xtanx.tex }

\noindent
There is root between each vertical asymptote.
The first 10 positive roots satisfy
$$x_n\in \Big({(2n-1)\pi\over 2},{(2n+1)\pi\over 2}\Big) 
\qquad\hbox{where}\qquad n=1,2,\ldots,10.$$
Steffensen's method
requires the initial guess be very close to the correct root
to converge because of the aymptotes.
To overcome this problem we check whether each Steffensen's 
iterate remains in the interval where the $n$-th root is located.
If it does not remain in the interval, then we take a bisection 
step and try again.
The program is
\listing{stefbisect.c}
and the output is
\listout{stefbisect.out}
An alternative method is to invert the equations and work 
with $\arctan(y)$.  This is implemented in the program
\listing{stefatan.c}
which produces the output
\listout{stefatan.out}
It is worth noting that this second method involves transforming
the original problem into something easier to solve
and is significantly more efficient than the first.
\vfill\eject

\qn [Kincaid and Cheney Problem 3.5\#1] \ Use Horner's
algorithm to find $p(4)$ where
$$
    p(z)=3z^5-7z^4-5z^3+z^2-8z+2.
$$
\medskip\noindent
We arrange the calculation as
$$
\vbox{\halign{ \hfil $#$\quad $|$&&\quad\hfil $#$\cr
        &3&-7&-5&1&-8&2\cr
            4&&12&20&60&244&944\cr
\noalign{\vglue2pt\hrule\vglue2pt}
            &3&5&15&61&236&946\cr}}
$$
Therefore $p(4)=946$.
\vfill\eject

\qn [Kincaid and Cheney Problem 3.5\#5] \ For the polynomial
$$
    p(z)=3z^5-7z^4-5z^3+z^2-8z+2
$$
find a disk centered at the origin that contains all the roots.
\medskip\noindent
By the localization theorem all roots lie inside an open disk
centered at the origin of radius
$$\eqalign{
    \rho&=1+|a_n|^{-1}\max\big\{\, |a_k| : k=0,1,\ldots, n-1\,\big\}\cr
    &=1+3^{-1}\max\{ 2, 8, 1, 5, 7\}=1+8/3=11/3.
}$$
\vfill\eject

\qn [Kincaid and Cheney Problem 3.5\#19] \ Prove that if $p$ is
a polynomial of degree $n$ having real coefficients, then
$$
    (n-1)\big(p'(x)\big)^2\ge np(x)p''(x).$$
Assume that the roots are real.
\def\P{{\cal P}}%
\medskip\noindent
We prove this statement by induction.  Let $\P_n$ be the
set of all polynomials of degree $n$ having real coefficient and real
roots.
\smallskip
{\bf Base case:}  If $n=1$ then the statement reduces to the fact
that $0\ge 0$.
\smallskip
{\bf Induction step:}  Suppose
    $(n-1)\big(p_n'(x)\big)^2\ge n p_n(x)p_n''(x)$
for every $p_n\in\P_n$.  Let $p_{n+1}\in\P_{n+1}$.  Claim that
$n\big(p_{n+1}'(x)\big)^2\ge (n+1) p_{n+1}(x)p_{n+1}''(x).$

Let $r\in\R$ be a root of $p_{n+1}$.  Then there is $p_n\in\P_n$
such that
$$\eqalignno{
    p_{n+1}(x)&=(x-r)p_n(x)\cr
\noalign{\noindent and therefore}
    p_{n+1}'(x)&=p_n(x)+(x-r)p_n'(x)\cr
    p_{n+1}''(x)&=2p_n'(x)+(x-r)p_n''(x).\cr
}$$
For notational simplicity write $p_n$ in place of $p_n(x)$
and so forth.  Estimate using the induction hypothesis and
completing the square to obtain
$$\eqalign{
    (n+1)p_{n+1}p_{n+1}''
    &=(n+1)(x-r)p_n\big(2p_n'+(x-r)p_n''\big)\cr
    &=2(n+1)(x-r)p_n p_n' +(n+1)(x-r)^2p_n p_n''\cr
    &\le 2(n+1)(x-r)p_n p_n' +(n+1)(x-r)^2{n-1\over n}(p_n')^2\cr
    &=2(n+1)(x-r)p_n p_n' +{n^2-1\over n}(x-r)^2(p_n')^2\cr
    &=n\Big\{\big[(x-r)p_n'\big]^2+2p_n\big[(x-r)p_n'\big]\Big\}\cr
        &\qquad\qquad\qquad
            -{1\over n}\Big\{\big[(x-r)p_n'\big]^2
            -2np_n\big[(x-r)p_n'\big]\Big\}\cr
    &=n\big((x-r)p_n'+p_n\big)^2-np_n^2
            -{1\over n}\big((x-r)p_n'-np_n\big)^2+np_n^2\cr
    &=n\big((x-r)p_n'+p_n\big)^2
            -{1\over n}\big((x-r)p_n'-np_n\big)^2\cr
    &\le n\big((x-r)p_n'+p_n\big)^2=n(p_{n+1}')^2.
}$$
This completes the induction.
\vfill\eject

\qn [Kincaid and Cheney Computer Problem 3.5\#5] \ Write a computer routine
that uses Newton's method and deflation for finding all roots of
a polynomial.  Test the routine using Wilkinson's polynomial
$$
	p_{20}(x)=\prod_{n=1}^{20} (x-n).
$$
\medskip\noindent
Using Newton's method to find a root of an arbitrary polynomial
requires care in choosing the initial guess $x_0$.  In the code
that follows we use Theorem 3 in Kincaid and Cheney page 110 on 
the localization of polynomial roots to guide choices for the 
initial guess and to determine whether Newton's iteration has
diverged.   

Let $\rho_{\rm max}$ be $\phi$ times the bound obtained by 
Theorem 3 where $\phi=(1+\sqrt 5)/2$
and $\rho_n=\mu \phi^n$ where $\mu$ is the unit roundoff error.
Choose $N$ to be the largest integer such that $\rho_N\le \rho_{\rm max}$.
The following heuristics are used to guess $x_0$.
Since it is desired to find roots of smallest magnitude first,
we always try $x_0=0$ as the first guess.  If this doesn't work
we guess randomly an $x_0$ inside a circle centered $\rho_n$
where $n=0,1,\ldots,N$ until a root is found.

It is worth noting our strategy for guessing $x_0$
is simple, but may not lead to a root in all cases.  If a certain
maximum number of initial guesses $K$ has been tried without
finding a root the program exits with an error.

The program{\tt\ roots.c }is given by
\listing{roots.c}
and Wilkinson's polynomial{\tt\ wilkpoly.dat }is
given by
\listing{wilkpoly.dat}
Running the program with the command
\medskip
\beginverbatimb
	$ ./roots <wilkpoly.dat
\endverbatim
\medskip\noindent
produces the output
\listout{roots.out}
The roots are sorted and the polynomial is deflated in
order of roots in increasing magnitude to reduce rounding error.
To further reduce rounding error, each root
obtained using Newton's method on a reduced polynomial has
been refined by applying Newton's method to the original
polynomial.
However, note that although the{\tt\ long double complex }data type
used in this program retains over 18 digits of precision the
resulting roots are accurate to only 7 significant digits.
This is because the conditioning number corresponding to the
problem of finding the roots of a polynomial is very large.
\vfill\eject

\qn [Kincaid and Cheney Problem 3.6\#1] \ Solve the
system of equations
$$
    \cases{ x-2y+y^2+y^3-4=0\cr
        -x-y+2y^2-1=0}
$$
by the homotopy method starting with the point $(x_0,y_0)=(0,0)$.
Perform all calculations without recourse to numerical methods.
\medskip\noindent
Following Kincaid and Cheney Example 2 on page 133 we write
$$\eqalign{
    h(t,x,y)&=tf(x,y)+(1-t)\big(f(x,y)-f(x_0,y_0)\big)\cr
        &=f(x,y)+(1-t)f(0,0)
    =
        \left[
        \matrix{x-2y+y^2+y^3-8-4t\cr -x-y+2y^2-2-t}
    \right].
}$$
Therefore
$$
    {dh\over ds}=Dh {d\over ds}\left[\matrix{t\cr x\cr y}\right]
        =\left[\matrix{
            -4&1&3y^2+2y-2\cr
            -1&-1&4y-1\cr
        }\right]\left[\matrix{t'\cr x'\cr y'}\right]=0.
$$
We use Cramer's rule via Lemma 1 on page 135 to choose a
solution $\big(t(s),x(s),y(s)\big)$ to this equation that
is continuous in $s$.   We obtain the system
$$\eqalign{
    t'&=\det\left[\matrix{1&3y^2+2y-2\cr-1&4y-1}\right]
        =3y^2+6y-3\cr
    x'&=-\det\left[\matrix{-4&3y^2+2y-2\cr-1&4y-1}\right]
        =-3y^2+14y-2\cr
    y'&=\det\left[\matrix{-4&1\cr-1&-1}\right]
        =5\cr
}$$
This system of ordinary differential equations is triangular
and solved by integrating first for $y$, then for $x$ and
finally for $t$.  The computation is
$$\eqalign{
    y(s)&=y_0+\int_0^s 5\,du= 5s\cr
    x(s)&=x_0+\int_0^s \big(-3(5u)^2+14(5u)-2\big)du
        =-25s^3+35s^2-2s\cr
    t(s)&=\int_0^s \big(3(5u)^2+6(5u)-3\big)du
        =25s^3+15s^2-3s\cr
}$$
Now solve for $s$ such that $t(s)=1$.  We obtain
$$
    25s^3+15s^2-3s-1=0$$
which we first try to factor over ${\bf Q}$.  By the rational root
theorem the possible roots are
$$
    \Big\{ \pm 1,\pm {1\over 5}, \pm {1\over 25}\Big\}.$$
Testing these yields that $s_1=-1/5$ is a root so we obtain
$$(5s+1)(5s^2+2s-1)=0.$$
The remaining roots can be found by the quadratic formula as
$$
    s_2={-2+\sqrt{4+20}\over 10}={-1+\sqrt 6\over 5}
\qquad\hbox{and}\qquad
    s_3={-1-\sqrt 6\over 5}.
$$
The corresponding solutions are
$$\eqalign{
    \big(x(s_1),y(s_1)\big)&=(2,-1)\cr
    \big(x(s_2),y(s_2)\big)&
        =\big(14-5\sqrt 6,-1+\sqrt 6\,\big)\cr
    \big(x(s_3),y(s_3)\big)&
        =\big(14+5\sqrt 6,-1-\sqrt 6\,\big).\cr
}$$
We now check the solutions using Maple.
The Maple script is
\listing{prob361.mpl}
and the output is
\listout{prob361.out}
which shows we have indeed found three roots.
\vfill\eject

\qn [Kincaid and Cheney Problem 3.6\#2] \ Consider the
homotopy
$$h(t,x)=tf(x)+(1-t)g(x)$$ in which
$$
    f(x)=x^2-5x+6
\qquad\hbox{and}\qquad
    g(x)=x^2-1.
$$
Show that there is no real path connecting a root of $g$ to
a root of $f$.
\medskip\noindent
If there were such a path then it would have to pass through
a point when $t=1/2$ of the form $(1/2,x)$ where $h(1/2,x)=0$.  However
$$
    h(1/2,x)={1\over 2}(x^2-5x+6)+{1\over 2}(x^2-1)
        ={1\over 2}(2x^2-5x+5)=0
$$
implies by the quadratic formula that
$$
    x={5\pm\sqrt{25-40}\over 4}\not\in \R.$$
Therefore there is no real path connecting a roots of $g$ to
a root of $f$.
\vfill\eject

\end
