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Quadratic equations always have two
solutions (which may be coincident) but until we know about complex
numbers, we can't always find the solutions. This problem helps you
to track the solutions (or roots) of quadratic equations and to
learn about complex numbers. Don't be put off because the question
is long. This is because it is written for students who have never
met complex numbers before. Anyway short questions are often much
harder than long ones.
Throughout this
problem you are in control of the general quadratic equation $x^2 +
px + q = 0$. You can change this equation by moving the point $(p,
q)$ in the red frame and you have to investigate what happens to
the roots of this equation as you change the values of $p$ and
$q$.
The blue frame
shows the graph of $y=x^2 + px + q$.
Use either the
arrow keys or the mouse to move the spot in the red box.
This text is usually replaced by the Flash movie.
(1) What happens to the graph of $y=x^2 + px + q$ if you keep
$p$ constant and change $q$? What do you notice about the
intersections of the graph of $y=x^2 + px + q$ with the $x$ axis
for
(a) $p=-5,\ q=-6\quad $ (b) $p=-5,\ q=4\quad $ (c) $p= -5,\ q=
7$?
(2) What happens to the graph of $y=x^2 + px + q$ if you keep
$q$ constant and change $p$? What do you notice about the
intersections of the graph of $y=x^2 + px + q$ with the $x$ axis
for
(a) $p=-10,\ q=16\quad $ (b) $p=-8,\ q=16\quad $ (c) $p= -6,\
q=16$?
(3) What happens to the graph of $y = x^2 + px +q$ as you move
the point $(p,q)$ around in the red frame below? Identify two
different regions in the $(p,q)$ plane and explain the significance
of the two regions and the boundary between them. [To help you to
answer this question the point leaves a coloured track as you move
it around.]
This text is usually replaced by the Flash movie.
The green frame is called the
Argand Diagram and it shows points
with coordinates $(u,v)$ for variables $u$ and $v$. Look for the
two points in the Argand diagram that move as you change the
driving point $(p,q)$ in the red frame, and in doing so change the
quadratic equation and its roots, and you will see for yourself how
the Argand diagram shows the roots of the quadratic
equations.
(4) What can you say about the relationship between $p$ and $q$
when the the roots show up on the $u$-axis in the Argand Diagram
(called the
real axis
)?
What can you say about the values of $p$ and $q$ when the roots
show up on the $v$-axis in the Argand Diagram (called the
imaginary axis )?
(5) Set the point $(p,q)$ to $p=0, q=1$ and read the coordinates of
the points representing the two roots in the Argand diagram. The
corresponding quadratic equation is $x^2 + 1 = 0$.
We know that if we square a real number we always get a positive
result so the square root of a negative number is not a real
number. To solve such equations we have to broaden our horizons to
take in $\sqrt{-1}$ which mathematicians call $i$ and with it an
expanded set of numbers called complex numbers.
You have just found two complex numbers $i=(u,v)=(0,1)$ and
$-i=(u,v)=(0,-1)$ in the Argand diagram which satisfy the equation
$x^2 + 1 = 0$.
Real numbers are one dimensional and are represented by points
$(u,0)$ on the real axis in the Argand diagram. The set of real
numbers is contained in the bigger set of of complex numbers which
are 2-dimensional and are represented by points $(u,v)$ in the
Argand diagram. Equivalently the complex number $(u,v)$ can be
written as $u + iv$ where $i^2 = -1$ and in this form we can add,
subtract, multiply and divide complex numbers and they obey all the
laws of algebra that you have already learnt.
(6) Set the point $(p,q)$ to $p=-6, q=13$ and read the co-ordinates
of the points representing the two roots in the Argand Diagram. Now
use the quadratic formula to write down two solutions to the
equation $x^2 -6x +13 = 0$. The roots you have just found are
2-dimensional complex numbers and these roots involve the square
root of a negative number.
Write down the solutions $z_1$ and $z_2$ to $x^2-6x+13=0$ and,
using the ordinary rules of algebra and substituting $i^2 = -1$,
calculate
(a) $z_1 + z_2\quad $ (b) $z_1 \times z_2\quad $ (c) $z_1^2 - 6z_1
+13\quad $ and $\quad $ (d)$z_2^2 -6z_2 +13$.
(7) Prove that two complex roots of a quadratic equation $x^2 + px
+q = 0$ (where $p$ and $q$ are real numbers) are always given by
points in the Argand diagram that are reflections of each other in
the real axis and are of the form $z_1=u + iv$ and $z_2= u - iv$.
Prove further that the sum and product of the roots are both
real.
To find out more about complex numbers read the article
What are Complex Numbers?
NOTES AND BACKGROUND
In primary school we could find numbers to put in the boxes for
equations like $\square + 2 = 7$, $4\times \square = 8$ and
$\square \times \square = 9$ and we were really solving the
equations $ x + 2 = 7$, $4x = 8$ and $x^2 = 9$ without using
algebraic notation.
We could not solve equations like $x + 7 = 2$ until we learned
about negative numbers. We could not solve equations like $4x = 3$
until we learned about fractions (rational numbers) and we could
not solve equations like $x^2 = 3$ until we learned about square
roots and irrational numbers.
When we first learn to solve quadratic equations, by factorising,
completing the square or using the quadratic formula, we find that
some quadratic equations have two real roots, some quadratic
equations have a repeated root and some have no real roots. Surely
this is an unsatisfactory situation!
Just as our knowledge and understanding of numbers had to expand
for us to be able to solve all linear equations, so we have to
learn more about numbers in order to be able to solve all quadratic
equations.
Once we understand complex numbers, which are algebraically very
simple, then we can solve all
quadratic equations , that is we can find solutions, or
roots, of all equations of the form $x^2 + px + q = 0$ or,
equivalently, values of $x$ for which the function $F(x) = x^2 + px
+ q$ takes the value zero.
By the Fundamental Theorem of Algebra a polynomial equation $F(z) =
0$, where $F(z)$ is a polynomial of degree $n$, always has exactly
$n$ solutions, some of which may be repeated.