|
| From: | Riana Meyer |
| Subject: | FHSST-Maths: solving updated |
| Date: | Wed, 29 Jun 2005 17:15:17 +0100 |
| User-agent: | Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.6) Gecko/20040113 |
|
Hi Sam Please find attached the solving chapter with my additions. I am not exactly happy with how everything looks so I was just wondering if you could maybe give it a quick check and let me know if there is anything I can do to improve it. Since I am a Latex uber-novice I would appreciate any advice and thought it better to send it at this stage than struggle another 2 weeks and end up with the wrong thing anyway. So please let me know if I am on the right track at least. Some things I had issues with were: 1. Sub subsections - I want the rules headings to be bigger than the example headings and stand out more. Because, really, the examples are subsections of the rule sections. Jeezzz that is confusing but do you understand what I mean? 2. Rule 2 - It is the last two lines of page two and looks odd. it should be the beginning of page 3, how do I do that? 3. TIPS - I cant figure out a way for them to stand out more like in the word document. What environments do you suggest I use , the frame box doesn't seem to work? 4. eqn line 1.19 - is that the right way to do it or should I use \div ? Sorry for all the questions, but help will be appreciated since I would like to move one to the next bit now. Thanks Riana |
% Copyright (c) 2003 "Free High School Science Texts"
% This file is licensed under the terms of the GNU Free Documentation
% License, please see http://www.gnu.org/licenses/fdl.html
%===========================================================================%
\documentclass[10pt, a4paper, titlepage]{report}
\usepackage{color}
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\usepackage{amsmath, amsthm, amsfonts, amssymb}
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% so decimal commas look right
%\DeclareMathSymbol{,}{\mathord}{letters}{"3B}
%define two different macros to generate a comma
%one as an ordinary symbol and one as punctuation
%the comma is then made active in maths mode, and
%defined to look ahead to the next token
%if this token is a space, a punctuation comma is
%used else an ordinary comma is used. Thus 3,1415
%is formatted properly without a space and (2, 5) is
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{\catcode`\,=\active
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%the macro divrule#1#2 produces a rule the same width as
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\begin{document}
% FHSST-Maintainer: Sam Halliday <address@hidden>
% FHSST-State: in edit
% FHSST-Editor: Riana Meyer <address@hidden>
% FHSST-Start-Contributors:
% Andrea Prinsloo <address@hidden>
% - initial draft of quadratic equations (?)
% Sam Halliday
% - a little on quadratic factorising
% FHSST-Start-Contributors:
\chapter{Solving Equations}
\label{ms}
\section{Linear Equations}
\label{ms:l}
\begin{syllabus}
\item solve linear equations
\end{syllabus}
\subsection{Introduction}
Let's imagine you have a friend called Joseph. He picked up your test results
from the Biology class and now he refuses to tell what you scored, or what he
scored! Obviously you are trying everything to get him to tell you, and he
decides to tease you and makes you work it out for yourself. He says the
following:
\\{}
\newline
``I have 2 marks more than you and the sum of both our marks is equal to 14.
How much did we get?''
\\{}
\newline
Now if the numbers are simple like in the example, you might be able to work it
out in your head. Can you? But to make it easier, you can use a linear
equation!
\\{}
\newline
This is how it works:
\\{}
\newline
We use a placeholder for your amount and that placeholder is $x$. So:
\\{}
\newline
$You = x$
\\{}
\newline
Then we need a placeholder for Joseph
\\{}
\newline
$Joseph = y$
\\{}
\newline
BUT the trick is that we have some information about Joseph's mark, which is
that Joseph has 2 more than you. We need to use that, so how about:
\\{}
\newline
$Joseph = you + 2$
\newline
Or
\newline
$Joseph = (x + 2)$
\newline
Or
\newline
$y = (x + 2)$
\\{}
\newline
Now we need to use the last bit of information we have and that is:
\\{}
\newline
$You + Joseph = 14$
\\{}
\newline
Or using placeholders
\newline
$x + y = 14$
\\{}
\newline
Or substituting y
\newline
$x + (x+2) = 14$
\\{}
\newline
What we have here is the actual linear equation.
\\{}
\newline
You already know what an equation is but what does linear mean? Linear means
the highest power of the unknown variable, usually called $x$, is one.
\subsection{Solving Linear equations - the basics}
To find out what your test result is we need to now simplify this equation
until we only have the x on the one side of the equal sign and a value on the
other side. There are a few rules on how to simplify these equations to get a
value for x. They can be organized into 3 groups. Once we have worked through
them and we are sure about them, then we can attempt to find out what the
answer to our problem is. So here they are:
\subsubsection{Rule one - Addition or subtraction}
You are allowed to subtract or add any amount as long as you do it on both
sides of the equal sign:
\subsection*{\emph{Example 1:}}
\begin{eqnarray}
& & x + 5 = -6 \\
& \Rightarrow\ & x + 5 - 5 = -6 - 5 \\
& \Rightarrow\ & x = -11
\end{eqnarray}
\subsection*{\emph{Example 2:}}
\begin{eqnarray}
& & x - \frac{1}{2} = 7 \\
& \Rightarrow\ & x - \frac{1}{2} + \frac {1}{2} = 7 + \frac{1}{2} \\
& \Rightarrow\ & x = \frac{15}{2}
\end{eqnarray}
\subsubsection{Rule two - Multiply or divide}
The same principle applies for multiplication and division:
\subsection*{\emph{Example 1:}}
\begin{eqnarray}
& & 2x = 9 \\
& \Rightarrow\ & \frac{2x}{2} = \frac{9}{2} \\
& \Rightarrow\ & x = \frac{9}{2}
\end{eqnarray}
\subsection*{\emph{Example 2:}}
\begin{eqnarray}
& & \frac{x}{4} = 5 \\
& \Rightarrow\ & \frac{x}{4} \times 4 = 5 \times 4 \\
& \Rightarrow\ & x = 20
\end{eqnarray}
\subsubsection{Rule three - Fractions}
If $x$ is multiplied by a fraction we need to divide both sides of the equal
sign with that fraction to get $x$ alone. We do that by flipping the fraction
around and then multiplying both sides with it
\subsection*{\emph{Example 1:}}
\begin{eqnarray}
& & (\frac{3}{2})x = 7 \\
& \Rightarrow\ & (\frac{3}{2})x(\frac{2}{3}) = 7(\frac{2}{3}) \\
& \Rightarrow\ & x = \frac{14}{3}
\end{eqnarray}
These are the basic rules to apply when simplifying a linear equation. But most
linear equations will require a few combinations of these before $x$ is sitting
alone on the one side of the equal sign. That means we might have to make use
of the rules above a number of times one after the other. Let's do a few
examples where we will use multiple steps to solve the equation:
\subsection{Solving linear equations - Combining the basics in a few steps}
\framebox(450,20)[c]{TIP: Start with eliminating the terms without x, that way
you avoid having to calculate too many fractions}
\subsection*{\emph{Example 1:}}
\begin{eqnarray}
& & 7 + 5x = 62\\
& \Rightarrow\ & 7 + 5x - 7 = 62 -7 \\
& \Rightarrow\ & 5x = 55 \\
& \Rightarrow\ & \frac{5x}{5} = \frac{55}{5} \\
& \Rightarrow\ & x = 11
\end{eqnarray}
\subsection*{\emph{Example 2:}}
\begin{eqnarray}
& & 55 = 5x + \frac{3}{4}\\
& \Rightarrow\ & 55 - \frac{3}{4} = 5x + \frac{3}{4} - \frac{3}{4}\\
& \Rightarrow\ & 54(\frac{1}{4}) = 5x\\
& \Rightarrow\ & \frac{217}{4} = 5x \\
& \Rightarrow\ & \frac{217}{4} \times \frac{1}{5} = 5x \times \frac{1}{5}
\end{eqnarray}
Doing that is the the same as dividing by 5\\
\begin{equation}
& \Rightarrow\ & \frac{217}{20} = x
\end{equation}
\begin{quotation} TIP: Start by moving all the terms with x to the one side and
all the terms without x to the opposite side of the equal sign. Remember we can
do that by changing the sign of the term
\end{quotation}
\subsection*{\emph{Example 3:}}
\begin{eqnarray}
& & 5x = 3x + 45 \\
& \Rightarrow\ & 5x - 3x = 45\\
& \Rightarrow\ & 2x = 45\\
& \Rightarrow\ & \frac{2x}{2} = \frac{45}{2}\\
& \Rightarrow\ & x = 22\frac{1}{2}
\end{eqnarray}
\subsection*{\emph{Example 4:}}
\begin{eqnarray}
& & 23x - 12 = 6 + 2x\\
& \Rightarrow\ & 23x - 2x -12 = 6\\
& \Rightarrow\ & 23x - 2x = 6 + 12\\
& \Rightarrow\ & 21x = 18\\
& \Rightarrow\ & x = \frac{21}{18}
\end{eqnarray}
\subsection*{\emph{Example 5:}}
\begin{eqnarray}
& & 12 - 6x + 34x = 2x - 24 - 64\\
& \Rightarrow\ & -6x + 34x = 2x -24 - 64 - 12\\
& \Rightarrow\ & -6x + 34x - 2x = -24 - 64 - 12\\
& \Rightarrow\ & 26x = -100\\
& \Rightarrow\ & x = - \frac{100}{26}\\
& \Rightarrow\ & x = - \frac{50}{13}
\end{eqnarray}
We simplified the answer - but this is not necessarily a required step
\\
\newline
\framebox{TIP: If there are parentheses (brackets) in the equation, start by
removing them - multiply with the coefficient}
\subsection*{\emph{Example 6:}}
\begin{eqnarray}
& & -3(3x - 4) = 8\\
& \Rightarrow\ & -9x + 12 = 8\\
& \Rightarrow\ & -9x = 8 - 12\\
& \Rightarrow\ & -9x = -4\\
& \Rightarrow\ & x = \frac{4}{9}
\end{eqnarray}
see the term is now positive - do you remember why?
\subsection*{\emph{Example 7:}}
\begin{equation}
& & 6x + 3x = 4 - 5(2x - 3)
\end{equation}
lets start with the parentheses - don't forget the minus!\\
\begin{equation}
& \Rightarrow\ & 6x + 3x = 4 -10x + 15\\
\end{equation}
next we move the like terms to their own sides\\
\begin{eqnarray}
& \Rightarrow\ & 6x + 3x + 10x = 4 + 15\\
& \Rightarrow\ & 19x = 19 \\
& \Rightarrow\ & x = 1
\end{eqnarray}
\subsection*{\emph{Example 8:}}
\begin{equation}
& & 8(3x - 14) - 34 = 2(4x - 22) - 5(3 + 2x)
\end{equation}
Looks like a big one? Lets take it step by step\\
\begin{equation}
& \Rightarrow\ & 24x - 112 - 34 = 8x - 44 - 15 - 10x\\
\end{equation}
that's all the brackets gone\\
\begin{eqnarray}
& \Rightarrow\ & 24x -112 - 34 -8x +10x = -44 - 15\\
& \Rightarrow\ & 24x - 8x + 10x = -44 - 15 + 112 + 34
\end{eqnarray}
and now solve!\\
\begin{eqnarray}
& \Rightarrow\ & 26x = 87\\
& \Rightarrow\ & x = \frac{87}{26}
\end{eqnarray}
And that is it for our examples. This covers all the types of linear equations
you can be expected to solve. It's the best to always keep your priority of
steps in mind and then just simply do them one by one. If you are unsure about
your answer you can just substitute it into your original equation and see if
you get the same value for both sides:
\subsection*{\emph{Example :}}
\begin{eqnarray}
& & 5(x - 3) = 5\\
& \Rightarrow\ & 5x - 15 = 5\\
& \Rightarrow\ & 5x = 5 + 15\\
& \Rightarrow\ & 5x = 20 \\
& \Rightarrow\ & x = 4
\end{eqnarray}
Test:
\begin{eqnarray}
& & 5(4 - 3) = 5\\
& \Rightarrow\ & 5(1) = 5\\
& \Rightarrow\ & 5 = 5
\end{eqnarray}
and there we see it works!\\
\newline
Now lets get back to our original problem of your test results! The linear
equation was:
\begin{eqnarray}
& & x + (x+2) = 14\\
& \Rightarrow\ & x + x + 2 = 14\\
& \Rightarrow\ & x + x = 14 - 2\\
& \Rightarrow\ & 2x = 12\\
& \Rightarrow\ & x = 6
\end{eqnarray}
You scored 6 for your Biology test and Joseph scored 6 +2 = 8!
\section{Quadratic Equations}
\label{ms:q}
\begin{syllabus}
\item solve quadratic equations by factorisation, completing the square and
quadratic formula
\item identify ``not real'' numbers and how they occur. (see \ref{mp:s})
\end{syllabus}
\subsection{The Quadratic Function}
\nts{these notes have just been copied and pasted from the older structure notes
and have not been written to the syllabus. it needs a serious edit and the
worked example methodology needs redone (we now do inline examples and
analogies with worked examples and exercises at the end of the chapter.)} A
\emph{quadratic} or \emph{parabolic function} is a function of the form $f(x) =
ax^{2} + bx + c$.
\nts{very quick notes by sam for simple quadratics... this would be best as a
decision tree. make the student appreciate that its basically trial and error
to get the answer, but you can do some detective work first to eliminate most
possibilities
\begin{itemize}
\item write the problem in the form $ax^{2} + bx + c = 0$ (with $a$ positive)
\item write down two brackets, with an $x$ in each, leaving room for a number on
each side
\begin{equation}
\label{eq:ms:q:steps:1}
(\quad x\qquad)(\quad x\qquad)
\end{equation}
\item write out your options (in a table at the side) for multiplying two
numbers together to give $a$. these numbers should go in front of the $x$s in
your brackets.
\item if $c$ is positive, then the other two numbers you need are either both
positive or both negative. they are both negative if $b$ is negative, and both
positive id $b$ is positive. if $c$ is negative, it means only one of your
numbers is negative, the other one beng positive.
\item your two numbers should multiply to give $c$, so write out your options
(in a table off to the side). if each number is multiplied by the number in
front of the $x$ in the other bracket, then added together it gives you $b$.
so try different combinations of the numbers you have written). if it doesn't
work, go back to the 3rd step and try a different combination of numbers to
give you $a$.
\item once you get an answer, multiply out your brackets again just to make sure
it works (sanity check).
\end{itemize}
damn thats long winded!}
\subsection*{\emph{Worked Example:}}
Q: Draw a graph of the quadratic function $y = x^{2} - x - 6$.
A: First let us set up a table of $x$ and $y$ values:
\begin{table}[!htb]
\begin{center}
\centering
\begin{tabular}{cccccccccccc}
\hline
$x:$ & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
$y:$ & 24 & 14 & 6 & 0 & -4 & -6 & -6 & -4 & 0 & 6 & 14 \\
\hline
\end{tabular}
\end{center}
\end{table}
The graph of this function is shown in figure \ref{fig:hsa:qf:ex1}.
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-6,-2)(6,6)
% \psset{yunit=0.2}
% \psaxes[Dy=5]{<->}(0,0)(-6,-10)(6,30)
%
\psdots(-5,24)(-4,14)(-3,6)(-2,0)(-1,-4)(0,-6)(1,-6)(2,-4)(3,0)(4,6)(5,14)
% \psplot[plotstyle=curve]{-5.1}{6}{x 2 exp x sub 6 sub}
% \end{pspicture}
% \caption{Graph of $y=x^2-x-6$}
% \label{fig:hsa:qf:ex1}
% \end{center}
% \end{figure}
Notice that the function can also be written as $y = (x + 2)(x - 3)$. This shows
that the $x$-intercepts (where $y = 0$) are $x = -2$ and $x = 3$, which agrees
with the graph. The $y$-intercept (where $x = 0$) is at $y = -6$.
\subsection{Writing a quadratic function in the form $f(x) = a(x-p)^{2} + q$.}
Consider the general form of the quadratic function $y = ax^{2} + bx + c$. Now
adding and subtracting the same factor $\frac{b^{2}}{4a}$ from this expression
does not change anything. Therefore
\begin{equation}
y = ax^{2} + bx + \frac{b^{2}}{4a} - \frac{b^{2}}{4a} + c
\end{equation}
Taking out a factor of $a$ then gives
\begin{equation}
y = a(x^{2} + \frac{b}{a} + (\frac{b}{2a})^{2}) + c -
\frac{b^{2}}{4a}
\end{equation}
The expression in brackets is then a perfect square so that
\begin{eqnarray}
& y & = a(x + (\frac{b}{2a}))^{2} + c - \frac{b^{2}}{4a} \\
& & = a(x - (-\frac{b}{2a}))^{2} + (c - \frac{b^{2}}{4a})
\end{eqnarray}
which can be written in the form
\begin{equation}
y = a(x - p)^{2} + q
\end{equation}
where $p = -\frac{b}{2a}$ and $q = c - \frac{b^{2}}{4a}$.
Since $(x - p)^{2}$ is a perfect square and therefore always positive, $(x -
p)^{2}$ is at a minimum of 0 when $x = p$. This means that $y$ is minimum (if $a
> 0$) or maximum (if $a < 0$) when $x = p$ and $y = q$. This point $(p,q)$ is
therefore called the \emph{turning point}.
Now notice that the quadratic function is symmetric about $x = p$. In other
words $f(p + v) = f(p - v) = av^{2} + q$ for any real number $v$. This means
that the part of the quadratic to the right of the vertical line $x = p$ looks
like the part to the left of $x = p$ flipped about this line. Therefore we call
the line $x = p$ the \emph{axis of symmetry} of the parabola.
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-3,-1)(9,4)
% \psset{xunit=2,yunit=0.5}
% \psaxes[labels=none,ticks=none]{<->}(0,0)(-1.5,-2)(4.5,8)
% \psplot[plotstyle=curve]{-1}{4}{x 1.5 sub 2 exp}
% \psline[linestyle=dashed](1.5,-2)(1.5,8)
% \rput(1.8,-0.7){$x=p$}
% \psdots(-0.5,4)(3.5,4)
% \rput(-0.9,3.8){$f(p-v)$}
% \rput(3.9,3.8){$f(p+v)$}
% \psline[linestyle=dashed]{<->}(-0.5,4)(1.5,4)
% \psline[linestyle=dashed]{<->}(1.5,4)(3.5,4)
% \rput(0.5,4.7){$v$}
% \rput(2.5,4.7){$v$}
% \end{pspicture}
% \caption{Graph of TODO \nts{original BMP (qf-gen1.bmp) missing, i made
this
% up on the fly}}
% \label{fig:hsa:qf:gen1}
% \end{center}
% \end{figure}
\subsection*{\emph{Worked Example:}}
Q:~ Consider the quadratic function $f(x) = -x^{2} + 6x - 5$. Put this function
into the form $f(x) = a(x - p)^{2} + q$ and thus find the turning point and axis
of symmetry. Plot a graph of $f(x)$ showing all the intercepts.
A: ~ First we shall write the quadratic in the form $f(x) = a(x - p)^{2} + q$.
\begin{eqnarray}
& f(x) & = -x^{2} + 6x - 5 \\
& & = -(x^{2} - 6x + 5) \\
& & = -(x^{2} - 6x + 9 - 9 + 5) \\
& & = -(x^{2} - 6x + 9) + 4 \\
& & = -(x^{2} - 3)^{2} + 4
\end{eqnarray}
Therefore the turning point is (3,4) and the axis of symmetry is $x = 3$ (in
other words $p = 3$ and $q = 4$).
Now to plot the graph we need to know the intercepts. The $y$-intercept is $y =
f(0) = -5$. The $x$-intercepts can be found by solving the equation $f(x) = 0$
as follows:
\begin{eqnarray}
& & -x^{2} + 6x - 5 = 0 \\
& \Rightarrow & x^{2} - 6x + 5 = 0 \\
& \Rightarrow & (x - 1)(x - 5) = 0 \\
& \Rightarrow & x = 1 ~~ \textrm{or} ~~ x = 5
\end{eqnarray}
Note that the technique of writing $x^{2} - 6x + 5 = (x - 1)(x - 5)$ is called
factorisation. We shall learn more about this in the following chapter. For
now just check that this is true by multiplying out the brackets.
Thus the graph of the quadratic function $f(x)=-x^{2}+6x-5$ is
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-0.9,-0.5)(10.8,5)
% \psset{xunit=1.8}
% \psaxes{<->}(0,0)(-0.5,-0.5)(6,5)
% \psdots(1,0)(5,0)(3,4)
% \psplot[plotstyle=curve]{0.9}{5.1}{0 x 2 exp sub 6 x mul add 5 sub}
% \rput(0.9,0.6){$x_1$}
% \rput(5.1,0.6){$x_2$}
% \psline[linestyle=dashed](3,0)(3,5)
% \end{pspicture}
% \caption{Graph of TODO \nts{original BMP (qf-eg2.bmp) missing, i made this
% up on the fly}}
% \label{fig:hsa:qf:ex2}
% \end{center}
% \end{figure}
\subsection{What is a Quadratic Equation?}
An equation of the form $ax^{2} + bx + c = 0$ is called a \emph{quadratic
equation}. Solving this equation for $x$ is the same as finding the roots
($x$-intercepts) of the quadratic function $f(x) = ax^{2} + bx + c$.
\subsection{Factorisation}
We have already seen examples of solving for the roots of a quadratic function
by writing this function as the multiple of two brackets. For example, $x^{2} -
x - 6 = (x + 2)(x - 3)= 0$ means that either $x = -2$ or $x = 3$. This is
called \emph{factorising} the quadratic function.
Knowing how to factorise a quadratic takes some practice, but here some general
ideas which are useful.
\begin{itemize}
\item{First divide the entire equation by any common factor of the
coefficients,
so as to obtain an equation of the form $ax^{2} + bx + c = 0$ where $a$, $b$
and $c$ have no common factors. For example, $2x^{2} + 4x + 2 = 0$ can be
written
as $x^{2} + 2x + 1 = 0$ by dividing by $2$.}
\item{Now, if $ax^{2} + bx + c = (rx + s)(ux + v)$, then $sv = c$ and $ru =
a$.
Therefore, by finding all the factors of $a$ and $c$, one can try all the
combinations and see if there is one which gives the correct result for $b
= su + rv$. }
\item{Once writing the equation in the form $(rx + s)(ux + v) = 0$, it then
follows that the two solutions are $x = -\frac{s}{r}$ and $x =
-\frac{u}{v}$.}
\end{itemize}
\subsection*{\emph{Worked Examples:}}
\textbf{Example 1:}
Q: ~ Solve the equation $x^{2} + 3x - 4$.
A: ~ Since $a = 1$, if this equation can be factorised, it must have the form
\begin{equation}
x^{2} + 3x - 4 = (x + s)(x + v) = x^{2} + (s + v)x + sv
\end{equation}
Now, as $sv = -4$, we know that $s = -2$, $v = 2$ or $s = -1$, $v = 4$ or $s =
1$, $v = -4$ (excluding the options which just involve interchanging $s$ and
$v$, which makes no difference to the final answer).
Also $s + v = 3$, so $s = -1$ and $v = 4$ is the correct combination. Thus the
quadratic equation can be written as
\begin{equation}
x^{2} + 3x - 4 = (x - 1)(x + 4) = 0
\end{equation}
Therefore the solutions are $x = 1$ and $x = -4$.
\textbf{Example 2:}
Q: ~ Find the roots of the quadratic function $f(x) = -2x^{2} + 4x - 2$.
A: ~ We must find the solutions to the equation $f(x) = -2x^{2} + 4x - 2 = 0$.
First we divide both sides of the equation be a factor of $-2$. This gives the
equation
\begin{equation}
x^{2} - 2x + 1 = 0
\end{equation}
Now, let us assume that
\begin{equation}
x^{2} - 2x + 1 = (x + s)(x + v) = x^{2} + (s + v)x + sv
\end{equation}
Then $sv = 1$ and therefore either $s = v = 1$ or $s = v = -1$. Since $s + v =
-2$, it follows that $s = v = -1$ and thus
\begin{equation}
x^{2} - 2x + 1 = (x - 1)(x - 1) = (x - 1)^{2} = 0
\end{equation}
The only solution is therefore $x = 1$.
\textbf{Example 3:}
Q: ~ Solve the equation $2x^{2} - 5x - 12 = 0$.
A: ~ This equation has no common factors, but still has $a = 2$. Therefore, we
must look for a factorisation in the form
\begin{equation}
2x^{2} - 5x - 12 = (2x + s)(x + v) = 2x^{2} + (s + 2v)x + sv
\end{equation}
We see that $sv = -12$ and $s + 2v = -5$. All the options for $s$ and $v$ are
considered below.
\begin{table}[!htb]
\begin{center}
\centering
\begin{tabular}{ccc}
\hline
$s$ & $v$ & $s + 2v$ \\
\hline
2 & -6 & -10 \\
-2 & 6 & 10 \\
3 & -4 & -5 \\
-3 & 4 & 5 \\
4 & -3 & -2 \\
-4 & 3 & 2 \\
6 & -2 & 2 \\
-6 & 2 & -2 \\
\hline
\end{tabular}
\end{center}
\end{table}
\textbf{Note:} ~ Since we now have the factor of $2x$ in the first bracket, it
does make a difference, in this case, whether we interchange the $s$ and $v$
values. For example, $s = 2$, $v = -6$ and $s = -6$, $v = 2$ give different
solutions. We must therefore consider both options.
We can see that the combination $s = 3$ and $v = -4$ gives $s + 2v = -5$.
Therefore one can check that
\begin{equation}
2x^{2} - 5x + 12 = (2x + 3)(x - 4) = 0
\end{equation}
Therefore the solutions are $x = -\frac{3}{2}$ and $x = 4$.
\subsection{Completing the Square}
It is not always possible to factorize a quadratic function. We shall now derive
a general formula, which gives the solutions to any quadratic equation.
Consider a general quadratic equation $ax^{2} + bx + c = 0$.
Adding and subtracting $\frac{b^{2}}{4a}$ from the left-hand side does not
change the equation. Thus
\begin{equation}
ax^{2} + bx + \frac{b^{2}}{4a} - \frac{b^{2}}{4a} + c = 0
\end{equation}
Taking out a factor of $a$ from the 1st 3 terms gives
\begin{equation}
a(x^{2} + \frac{b}{a} + \frac{b^{2}}{4a^{2}}) - \frac{b^{2}}{4a} +
c = 0
\end{equation}
\begin{equation}
\Rightarrow a(x^{2} + \frac{b}{a} + (\frac{b}{2a})^{2}) =
\frac{b^{2}}{4a} - c
\end{equation}
We can now see that the term in brackets is the perfect square $(x +
\frac{b}{2a})^{2}$ and therefore
\begin{equation}
a(x + \frac{b}{2a})^{2} = \frac{b^{2}}{4a} - c
\end{equation}
Now dividing by $a$ and taking the square root of either side gives the
expression
\begin{equation}
x + \frac{b}{2a} = \pm\sqrt{\frac{b^{2}}{4a^{2}} - \frac{c}{a}}
\end{equation}
Finally, solving for $x$ implies that
\begin{equation}
x = -\frac{b}{2a} \pm\sqrt{\frac{b^{2}}{4a^{2}} - \frac{c}{a}}
= -\frac{b}{2a} \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}
\end{equation}
and taking the square root of $4a^{2}$ to obtain $2a$ gives
\begin{equation}
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\end{equation}
These are the solutions to the quadratic equation. Notice that there are two
solutions in general, but these may not always exists (depending on the sign of
the expression $b^{2} - 4ac$ under the square root).
\subsection*{\emph{Worked Examples:}}
\textbf{Example 1:}
\par{Q: ~ Solve for the roots of the function $f(x) = 2x^{2} + 3x - 7$.}
\par{A: ~ One should first try to factorise this expression, but in this
case it turns out that this is not possible. Therefore we must
make use of the general formula as follows:}
\begin{eqnarray}
& x & = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\
& & = \frac{-(3) \pm \sqrt{(3)^{2} -4(2)(-7)}}{2(2)} \\
& & = \frac{-3 \pm \sqrt{56}}{4} \\
& & = \frac{-3 \pm 2\sqrt{14}}{4}
\end{eqnarray}
\par{Therefore the two roots of the quadratic function are
$x = \frac{-3 + 2\sqrt{14}}{4}$ and $\frac{-3 - 2\sqrt{14}}{4}$.}
\bigskip
\textbf{Example 2:}
\par{Q: ~ Solve for the solutions to the quadratic equation $x^{2} - 5x + 8$.}
\par{A: ~ Again it is not possible to factorise this equation. The
general formula shows that}
\begin{eqnarray}
& x & = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\
& & = \frac{-(-5) \pm \sqrt{(-5)^{2} - 4(1)(8)}}{2(1)} \\
& & = \frac{5 \pm \sqrt{-7}}{2} \\
\end{eqnarray}
\par{Since the expression under the square root is negative these are not
real solutions ($\sqrt{-7}$ is not a real number). Therefore
there are no real solutions to the quadratic equation $x^{2} - 5x
+ 8$. This means that the quadratic function $f(x) = x^{2} - 5x +
8$ has no $x$-intercepts, but the entire function lies above the
$x$-axis.}
\par{\emph{Note to self: maybe add quadratic example about distance, velocity
and
acceleration ... object falling under action of gravity (giving
formula for distance as a function of time)?.}}
\subsection{Theory of Quadratic Equations}
\subsubsection{What is the Discriminant of a Quadratic Equation?}
\par{Consider a general quadratic function of the form $f(x) = ax^{2} + bx + c$.
The \emph{discriminant} is defined as $\Delta = b^{2} - 4ac$. This
is the expression under the square root in the formula for the
roots of this function. We have already seen that whether the
roots exist or not depends on whether this factor $\Delta$ is
negative or positive.
\subsubsection{The Nature of the Roots}
\subsubsection*{Real Roots:}
\par{Consider $\Delta \geq 0$ for some quadratic function $f(x) = ax^{2} + bx +
c$.
In this case there are solutions to the equation $f(x) = 0$ given
by the formula}
\begin{equation}
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}
\end{equation}
\par{since the square roots exists (the expression under the square root is
non-negative.)
These are the roots of the function $f(x)$.}
\par{There are various possibilities:}
\bigskip
\par{\emph{Equal Roots:}}
\par{If $\Delta = 0$, then the roots are equal and, from the formula, these
are given by }
\begin{equation}
x = -\frac{b}{2a}
\end{equation}
\bigskip
\par{\emph{Unequal Roots:}}
\par{There will be 2 unequal roots if $\Delta > 0$. The roots of $f(x)$
are \textbf{rational} if $\Delta$ is a perfect square (a number
which is the square of a rational number), since, in this case,
$\sqrt{\Delta}$ is rational.} Otherwise, if $\Delta$ is not a
perfect square, then the roots are \textbf{irrational}.}
\subsubsection*{Imaginary Roots:}
\par{If $\Delta < 0$, then the solution to $f(x) = ax^{2} + bx + c = 0$
contains the square root of a negative number and therefore there
are no real solutions. We therefore say that the roots of $f(x)$
are \emph{imaginary} (the function $f(x)$ does not intersect the
$x$-axis).}
\subsubsection*{Summary of Cases:}
\begin{itemize}
\item{Real Roots ($\Delta \geq 0$)}
\begin{itemize}
\item{Equal Roots ($\Delta = 0$)}
\item{Unequal Roots ($\Delta > 0$)}
\begin{itemize}
\item{Rational Roots ($\Delta$ a perfect square)}
\item{Irrational Roots ($\Delta$ not a perfect square)}
\end{itemize}
\end{itemize}
\item{Imaginary Roots ($\Delta < 0$)}
\end{itemize}
\par{\emph{Note to self: maybe add pictures showing these cases graphically?}}
\subsubsection*{\emph{Worked Examples:}}
\par{\textbf{Example 1:}}
\par{Q: ~ Consider the function $f(x) = 2x^{2} + 5x - 11$. Without solving
the equation $f(x) = 0$, discuss the nature of the roots of
$f(x)$.}
\par{A: ~ We need to calculate and classify $\Delta = b^{2} - 4ac$ according to
the cases for the roots.}
\begin{equation}
\Delta = (5)^{2} - 4(2)(-11) = 25 + 88 = 113
\end{equation}
\par{Now $\Delta$ is positive, so the roots are real and unequal. Also, since
113
is not a perfect square, the roots are irrational.}
\bigskip
\par{\textbf{Example 2:}}
\par{Q: ~ Consider the quadratic function $f(x) = x^{2} + bx + (2b - 5)$,
where $b$ is some constant. Classify the roots of this function as
far as possible.}
\par{A: ~ Let us calculate the discriminant}
\begin{equation}
\Delta = b^{2} - 4(1)(2b - 5) = b^{2} - 8b + 20
\end{equation}
\par{We shall now use a useful trick, which is to write the above
expression as a perfect square plus a number.}
\begin{equation}
\Delta = b^{2} - 8b + 20 = (b^{2} - 8b + 16) + 4 = (b - 4)^{2} + 4
\end{equation}
\par{Now $(b - 4)^{2} \geq 0$ because this is a perfect square. Therefore
we know that $\Delta \geq 4 > 0$.}
\par{We can thus say that $f(x)$ has real unequal roots. We do not know whether
$\Delta$ is a perfect square, since we do not know that value of
the constant $b$, and therefore we cannot say whether the roots
are rational or irrational.}
\section{Cubic Equations}
\label{ms:c}
\begin{syllabus}
\item (grade 12) solve cubic equations using factor theorem ``and other
techniques'' \nts{SH: i want to hit whoever wrote this syllabus}
\end{syllabus}
\section{Exponential Equations}
\label{ms:e}
\begin{syllabus}
\item solve exponential equations
\item (grade 12) switch between log and exp form of an equation
\end{syllabus}
\section{Trigonometric Equations}
\label{ms:t}
\begin{syllabus}
\item solve trig equations \nts{this implies that this section comes after trig,
which is also probably after geometry}
\end{syllabus}
\section{Simultaneous Equations}
\label{ms:s}
\begin{syllabus}
\item solve simultaneous equations algebraically and graphically
\end{syllabus}
\section{Inequalities}
\label{ms:i}
\begin{syllabus}
\item solve linear inequalities in 1 and 2 variables and illustrate
graphically
\item solve quadratic inequalities in 1 variable and illustrate graphically
\end{syllabus}
\subsection{Linear Inequalities}
Let us say that we are given a general inequality as follows:
\begin{equation}
ax + by + c \geq 0, ~~~~ ax + by + c > 0, ~~~~ ax + by + c \leq 0
~~~~ \textrm{or} ~~~~ ax + by + c < 0
\end{equation}
Now there are many possible values of $x$ and $y$, for which this may be true
(these will depend on the values of the constants $a$, $b$ and $c$). The set of
all the $(x,y)$ values which satisfy this inequality is called the
\emph{solution set}.
We shall now see how to draw the solution set on a graph. Let's consider the
following example.
\subsection*{\emph{Worked Example 1:}}
\par{Q: Find the solution set of the inequality $2x + y - 3 \geq 0$. }
\par{A: First we solve for $y$ by writing the inequality as}
\begin{equation}
y \geq -2x + 3
\end{equation}
\par{Now the function $y = -2x + 3$ is a straight line and the points $(x,y)$,
which satisfy the inequality are therefore all the points above
the line. These can be drawn as}
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-1,-1)(5,4)
% \pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]
% (-0.5,4)(2,-1)(5,-1)(5,4)
% \psaxes{<->}(0,0)(-1,-1)(5,4)
% \psline(-0.5,4)(2,-1)
% \psdots(0,3)(1.5,0)
% \rput(1.6,0.4){$\frac 32$}
% \end{pspicture}
% \caption{Graph of $y = -2x + 3$. Points satisfying $y\geq -2x+3$ are
% shaded}
% \label{fig:hsa:lp:ex1}
% \end{center}
% \end{figure}
\par{The shaded section on the graph, which shows the solution set, is
called the \emph{feasible region}.}
\par{Now sometimes $x$ and $y$ must satisfy more than one inequality.
In this case, we consider each inequality separately and then the
feasible region is where the feasible regions of each inequality
overlap. }
\subsection*{\emph{Worked Example 2:}}
\par{Q: Graphically represent the solution set for the following inequalities:}
\begin{eqnarray}
y\geq 1 \\
-2x + y - 5 < 0 \\
x + y - 10 \leq 0
\end{eqnarray}
\par{A: Solving for $y$ gives}
\begin{eqnarray}
& y & \geq 1 \\
& y & < 2x + 5 \\
& y & \leq -x + 10
\end{eqnarray}
Now we draw the solution set to each of the inequalities separately and find the
region where these overlap as shown below. We draw the line $y = 2x + 5$ as a
dashed line because the inequality is $<$ and not $\leq$ (the line is not
included in the feasible region).
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-2,-0.5)(6,5.5)
% \psset{xunit=0.5,yunit=0.5}
% \pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]
% (-2,1)(9,1)(1.667,8.333)
% \psaxes[Dx=2,Dy=2]{<->}(0,0)(-4,-1)(12,11)
% \psdots(-2.5,0)(0,1)(0,5)(0,10)(10,0)
% \psline(-4,1)(12,1)
% \psline(-1,11)(11,-1)
% \psline[linestyle=dashed](-3,-1)(3,11)
% \end{pspicture}
% \caption{Graph of TODO}
% \label{fig:hsa:lp:ex2}
% \end{center}
% \end{figure}
\subsection{What is a Quadratic Inequality}
\par{A \emph{quadratic inequality} is an inequality of the form
$ax^{2} + bx + c > 0$, $ax^{2} + bx + c \geq 0$, $ax^{2} + bx + c
< 0$ or $ax^{2} + bx + c \leq 0$.}
\subsection{Solving Quadratic Inequalities}
\par{Solving a quadratic inequality corresponds to working out in what
region a quadratic function lies above or below the $x$-axis. Here
are some examples showing how this is done.}
\subsection*{Worked Examples:}
\textbf{Example 1:}
\par{Q: ~ Find all the solutions to the inequality $x^{2} - 5x + 6 \geq 0$.}
\par{A: ~ Consider the function $f(x) = x^{2} - 5x + 6$. We need to find out
where $f(x) \geq 0$; in other words, where the function $f(x)$
lies above/on the $x$-axis.}
\par{We shall first work out where $f(x)$ intersects the $x$-axis by solving
the equation}
\begin{equation}
x^{2} - 5x + 6 = 0
\end{equation}
\par{which can be factorised to give}
\begin{equation}
(x - 3)(x - 2) = 0
\end{equation}
\par{The $x$-intercepts are therefore $x_1 = 2$ and $x_2 = 3$.}
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-2,-0.5)(8,4.5)
% \psset{xunit=2,yunit=0.5}
% \psaxes[Dy=2]{<->}(0,0)(-1,-1)(4,9)
% \psdots(2,0)(3,0)
% \psplot[plotstyle=curve]{-0.5}{4}{x 2 exp 5 x mul sub 6 add}
% \rput(2,0.7){$x_1$}
% \rput(3,0.7){$x_2$}
% \end{pspicture}
% \caption{Graph of $f(x)=x^2-5x+6$}
% \label{fig:hsa:qi:ex1}
% \end{center}
% \end{figure}
\par{We can see from figure \ref{fig:hsa:qi:ex1} that $f(x)$ is above/on the
$x$-axis
when $x \geq 3$ or $x \leq 2$.}
\par{Therefore the solution to the quadratic inequality is
$\{ x: x \geq 3$ or $x \leq 2 \}$ or in interval notation
$(-\infty,2] \cup [3,\infty)$.}
\par{\textbf{Note:} ~ The $x$-intercepts are included in this solution, since
the $f(x) \leq 0$ inequality includes the solution $f(x) = 0$. }
\bigskip
\textbf{Example 2:}
\par{Q: ~ Solve the quadratic inequality $-x^{2} - 3x + 5 > 0$.}
\par{A: ~ Let $f(x) = -x^{2} - 3x + 5$. The $x$-intercepts are solutions to
the quadratic equation }
\begin{eqnarray}
-x^{2}- 3x + 5 &=& 0 \\
\Rightarrow x^{2} + 3x - 5 &=& 0
\end{eqnarray}
\par{which has solutions (using the formula for the roots of a quadratic
function \nts{reference that equation}) given by}
\begin{eqnarray}
x &=& \frac{-3 \pm \sqrt{(3)^{2} - 4(1)(-5)}}{2(1)} \\
&=& \frac{-3 \pm \sqrt{29}}{2} \\
x_1 &=& \frac{-3 - \sqrt{29}}{2} \\
x_2 &=& \frac{-3 + \sqrt{29}}{2}
\end{eqnarray}
\par{The graph of $f(x)$ is shown in figure \ref{fig:hsa:qi:ex2}. The points
$x_1$
and $x_2$ (where the function $f(x)$ cuts the $x$ axis) are labelled.}
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-7.5,-1)(3,4)
% \psset{xunit=1.5,yunit=0.5}
% \psaxes[Dy=2]{<->}(0,0)(-5,-2)(2,8)
% \psdots(0,5)(-4.193,0)(1.193,0)
% \psplot[plotstyle=curve]{-4.5}{1.5}{5 x 2 exp sub 3 x mul sub}
% \rput(-4.3,0.7){$x_1$}
% \rput(1.3,0.7){$x_2$}
% \end{pspicture}
% \caption{Graph of $f(x)=-x^2-3x+5$}
% \label{fig:hsa:qi:ex2}
% \end{center}
% \end{figure}
\par{Now $f(x) > 0$ (the function is above the $x$-axis) when
$\frac{-3 - \sqrt{29}}{2} < x < \frac{-3 + \sqrt{29}}{2}$. }
\par{Therefore the solution to the inequality is
$\{x: \frac{-3 - \sqrt{29}}{2} < x < \frac{-3 + \sqrt{29}}{2}
\}$, or in interval notation $(\frac{-3 - \sqrt{29}}{2},\frac{-3 +
\sqrt{29}}{2})$.}
\par{\textbf{Note:} ~ The $x$-intercepts are not included in the solution
because the $>$ sign has been used and therefore $f(x) = 0$ does
not define a solution to the inequality.}
\bigskip
\textbf{Example 3:}
\par{Q: ~ Solve the inequality $4x^{2} - 4x + 1 \leq 0$.}
\par{A: ~ Let $f(x) = 4x^{2} - 4x + 1$. Factorising this quadratic function
gives $f(x) = (2x - 1)^{2}$, which shows that $f(x) = 0$ only when
$x = \frac{1}{2}$.}
% \begin{figure}[!ht]
% \begin{center}
% \begin{pspicture}(-3,-1.5)(9,6)
% \psset{xunit=6,yunit=3}
% \psaxes[Dx=0.5,Dy=0.5]{<->}(0,0)(-0.5,-0.5)(1.5,2)
% \psdots(0,1)(0.5,0)
% \psplot[plotstyle=curve]{-0.2}{1.2}{4 x 2 exp mul 4 x mul sub 1 add}
% \end{pspicture}
% \caption{Graph of $f(x)=4x^2-4x+1$}
% \label{fig:hsa:qi:ex3}
% \end{center}
% \end{figure}
\par{The function $f(x)$ lies below/on the $x$-axis only at the $x$-intercept.
Therefore the only solution to the inequality is $x =
\frac{1}{2}$. }
\section{Intersections}
\label{ms:in}
\begin{syllabus}
\item find solutions of 2 lines and interpret the common solution as the
intersection
\item find solutions of linear and quadratic, interpret the common solution(s)
as intersections
\end{syllabus}
\end{document}
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