Root Tracker

Stage: 5 Challenge Level: Challenge Level:1

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.
Full Screen Version
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.]

Full Screen Version
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.

search engine page

Published February 2007.