Polynomial Equations: Systematic Theory Summary, Challenging Examples and Problems

25 Pages·2017·0.658 MB·English

by  Demetrios P. Kanoussis

Polynomial Equations

A brief but systematic and comprehensive summary of
main ideas, techniques, methods and Theorems,
Challenging Examples and Problems,
All Problems are supplied with answers,
Ideal supplementary text for all Engineering, Math and
Physics students
Demetrios P. Kanoussis, Ph.D
2 DEMETRIOS P. KANOUSSIS
About the Author

Demetrios P. Kanoussis, Ph.D
Kalamos Attikis, Greece
[email protected]

Dr. Kanoussis is a professional Electrical Engineer and Mathematician. He received
his Ph.D degree in Engineering and his Master degree in Mathematics from Tennessee
Technological University, U.S.A, and his Bachelor degree in Electrical Engineering from
the National Technical University of Athens (N.T.U.A), Greece.
As a professional Electrical Engineer, Dr. Kanoussis has been actively involved in
the design and in the implementation of various projects, mainly in the area of the Inte-
grated Control Systems.
Regarding his teaching experience, Dr. Kanoussis has long teaching experience in
the field of Applied Mathematics and Electrical Engineering.
His original scientific research and contribution, in Mathematics and Electrical En-
gineering, is published in various, high impact international journals.
Additionally to his professional activities, teaching and research, Dr. Kanoussis is
the author of several textbooks in Electrical Engineering and Applied Mathematics. A
list of his publications is shown below:
Mathematics Textbooks
1) Polynomial Equations, e-book, May 2017.
2) Infinite Series and Products, e-book, April 2017.
3) Sequences of Real and Complex Numbers, e-book, March 2017.
4) Challenging Problems in Trigonometry, e-book, March 2015.
5) Topics in Applied Mathematics, paperback, November 2011,
(Greek Edition)
ALGEBRAIC EQUATIONS 3
Electrical Engineering Textbooks
1) The Direct Current Circuits Analysis, Vol. 2, e-book, May 2017.
2) Introduction to Electric Circuits Theory, Vol. 1, e-book, May 2017.
3) Introduction to Electric Circuits Theory, paperback, August 2013,
(Greek Edition)
4 DEMETRIOS P. KANOUSSIS
Polynomial Equations
Copyright 2017, Author: Demetrios P. Kanoussis.
All rights reserved. No part of this publication may be reproduced, distributed or
transmitted in any form or by any means, electronic or mechanical, without the prior
written permission of the author, except in the case of brief quotations and certain other
noncommercial uses permitted by copyright law.
Inquires should be addressed directly to the author,
Demetrios P. Kanoussis
[email protected]
This e book is licensed for your personal use only. This e book may not be resold or
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Thank you for respecting the work of this author.
First edition, May 2017
ALGEBRAIC EQUATIONS 5

Preface
Algebra traditionally deals with equations and systems of equations. The simplest
types of equations in Algebra are the so called polynomial equations, having the gen-
eral form, , where the coefficients

are either real or complex numbers.

The aim of this short book is to help the students to master some fundamental
techniques in solving polynomial equations using appropriate definitions, concepts and
theorems.
This book consists of three chapters.
The first chapter deals with first and second order equations, (Quadratic equa-
tions).
The second chapter deals with equations reducible to quadratic equations,
(Bi quadratic equations), or equations solved by means of an appropriate substitu-
tion. The method of substitution, in solving equations, is extremely powerful; however
there are no general rules as to which substitution is the proper one for each problem.
Substitution is a highly individual method of solution.
In the third chapter we state some general properties of polynomial equa-
tions, (The fundamental theorem of Algebra, proved rigorously for the first time by the
great C. F. Gauss in 1799, the Remainder Theorem, the Factor Theorem, and the com-
plex conjugate roots Theorem, the Rational Roots Theorem, etc.).
All solved examples and problems to be solved are carefully selected, in order to help
students to gradually acquire the necessary techniques, experience and computational
skills in problem solving.
All problems are supplied with answers.
Demetrios P. Kanoussis
6 DEMETRIOS P. KANOUSSIS
1. First and Second Degree (Quadratic) Equations
Polynomial Equations, constitute an important topic in Algebra, and have the gen-
eral form,
Px0 (1-1)
where Px is a polynomial of the variable x.
The coefficients of Px, in the general case, can be real or complex numbers.
Root or simply solution of Px0, is a number r (real or complex), such
that Pr0.
The number r is also known, as a root of the polynomial Px. Solving the equa-
tion Px0, means to find all the roots of the equation, i.e. all the roots of Px.
 If Px is a first degree polynomial, i.e. Pxaxb, where in general a and b
are complex numbers the corresponding equation is called a
first degree equation, and has the form axb0, the solution of which is easily
b
obtained to be x  .
a
So, for a first degree equation,

(1-2)

 The second degree equation, has the general form,
(1-3)
We shall, firstly, consider the case where a, b and c are real numbers, and we
shall provide a method to solve this equation.
 b c
Let Pxax2 bxcax2  x 
 a a
 b  b 2 c  b 2
ax2 2 x     
 2a 2a a 2a 
 b 2 b2 4ac
ax   .
 2a 4a2 
Therefore, the quadratic or second degree equation ax2 bxc0, is equivalent to
 b 2 b2 4ac
ax   0 (1-4)
 2a 4a2 
The quantity Δb2 4ac is known as the discriminant of the quadratic
equation. From eq. (1-4) it is obvious that,
1) If Δ 0, then the quadratic equation ax2 bxc0, has two real and distinct
roots, r and r , given by
1 2
ALGEBRAIC EQUATIONS 7
b Δ
r  , Δ0 (1-5)
1,2 2a
2) If Δ 0, then the quadratic equation ax2 bxc0, has a double real root r,
(i.e. in this case ), given by

b
r , Δ0 (1-6)
2a
3) If Δ0, then Δ 0, and the quadratic equation has two complex conjugate
roots, r and r , given by
1 2

) (1-7)

where of course i 1, is the imaginary unit i2 1.
Note 1: Formulas (1-5), (1-6) and (1-7), expressing the roots of the quadratic equa-
tions ax2 bxc0, in terms of the real coefficients a, b and c, are still valid in
case where a, b and c are complex numbers. Even in this case, the roots of
ax2 bxc0, with a, b and c complex, can be expressed as,
b b2 4ac
r  (1-8)
1,2 2a
as it is clear from Eq. (1-4), however, in this case the two roots r and r , in gen-
1 2
eral, will be complex numbers.

Note 2: In all cases, the roots r and r of the quadratic equation
1 2
ax2 bxc0, satisfy the following relations,
b c
r r   and r r  (1-9)
1 2 a 1 2 a
The proof of (1-9) is easy, and stems directly from the expressions of r and r , in
1 2
terms of the coefficients a, b, and c.

Example 1-1
Solve the quadratic equation x2 x10.

Solution
The discriminant Δ12 4113, so according to eq. (1-7),
1i 3 1i 3 1i 3
r  , i.e. r  and r  .
1,2 2 1 2 2 2
Example 1-2
2
1  1 
Solve for x the equation 3 1 2 1 10
   
x  x 
8 DEMETRIOS P. KANOUSSIS
Solution
1
If we set y 1, the given equation reduces to 3y2 2y10, having solution,
x
2 4431 2 16 24
y    , i.e.
1,2 23 6 6
1
y  and y 1.
1 3 2
1 1 3 1 1
From 1y  , we find x  , while from 1y 1, we find x  .
x 1 3 2 x 2 2
3 1
Therefore, the roots of the given equation, are x  and x 
1 2 2 2

Example 1-3
For which values of the parameter k, the quadratic equation kx2 2x9x2 2kx,
has a double root.

Solution
From kx2 2x9x2 2kxk1x2 21kx90, and this quadratic
equation will have a double root, if its discriminant Δ 0, i.e.
41k2 4k190k2 7k100. This is another quadratic equation for k,
having solution
7 494110 7 9 73
k    , i.e. k 5 or k 2.
1,2 2 2 2 1 2
For these two values of the parameter k, the given equation has a double root. For
3
example, for k5, we obtain, 4x2 12x9, having x  as a double root, while for
2
k2, we obtain x2 6x90, having x3 as a double root.

Example 1-4
Consider the quadratic equation ax2 bxc0, with real coefficients. Assuming that
this equation has two real roots r and r , and that a2  a2 abac , show that at
1 2
least one of the (real) roots, lies between 0 and 2.

Solution
b c
From a2  a2 abac  a2  a  abc  a  abc 1 1  (*)
a a
Taking into consideration eq. (1-9), (*) becomes,
1 1r r r r 1 1r r 1r  
1 2 1 2 1 2 1
1 1r 1r , i.e. 1r 1r 1 (**)
1 2 1 2
ALGEBRAIC EQUATIONS 9
However, if the product of two positive numbers is less than 1, then at least one of
them, say the 1r must be less than one, i.e. 0 1r 1, or 11r 1 or, equiva-
1 1 1
lently, 2r 0, i.e. 0r 2, Q.E.D.
1 1

Example 1-5
Let r and r be the two roots of the quadratic equation ax2 bxc0. Let also
1 2
S rn rn, where n is any positive integer, n2. Show that aS bS cS 0
n 1 2 n n1 n2

Solution
Since r and r are the roots of ax2 bxc0, we have,
1 2
ar2 br c0, and (*)
1 1
ar2 br c0 (**)
2 2
Multiplying (*) by rn2, (**) by rn2 and adding together the resulting equations,
1 2
we obtain arn rnbrn1 rn1crn2 rn20, or
1 2 1 2 1 2
aS bS cS 0, Q.E.D. (***)
n n1 n2

Note: Making use of (***), one may compute, successively S r3 r3, S r4 r4,
3 1 2 4 1 2
b
S r5 r5, ... etc. given that S r r  and
5 1 2 1 1 2 a
b2 c
S r2 r2 r r 2 2r r  2 .
2 1 2 1 2 1 2 a2 a
1 1 rn rn S
Also, one may easily compute the quantity Q    1 2 , i.e. Q  n .
n rn rn r r n n cn
1 2 1 2
 
a

Example 1-6
x y
If x ,y  and x ,y  are two solutions of the equation  , then show
1 1 2 2 y6x 3x8y
2
x y x y  16
that the quantity  2 1 1 2  will be either 0 or .
y y 9
 
1 2

Solution
First of all, we assume that x ,y ,x ,y are all different from 0, otherwise, the given
1 1 2 2
equation makes no sense.
10 DEMETRIOS P. KANOUSSIS
x y
From  , one easily obtains the equivalent equation,
y6x 3x8y
2
x x
3x2 2xyy2 0 or (since y0), 3  2 10. This equation, being
y y
x
quadratic for the ratio , will have two roots, say
 
y
x 2 4431 2 16 24
r  1    1, and
1 y 23 6 6
1
x 2 4431 2 16 24 1
r  2     .
2 y 23 6 6 3
2
2 2
x y x y  x x 
Now, the quantity  2 1 1 2   2  1  will be,
y y y y
   
1 2 2 1
x x x x 1
either 0, in case 2  1 r 1 or 2  1 r  ,
y y 1 y y 2 3
2 1 2 1
2 2
 1   4 16 x x
or  1    in the case 2  1.
 3   3 9 y y
2 1

PROBLEMS
1-1) For which values of the parameters p and q, the roots of the quadratic equation
x2 pxq0, are equal to p and q? (Assume pq0).
(Ans: p1, q2)

1-2) If k k 2l l , then prove that at least one of the equations
1 2 1 2
x2 k xl 0
1 1
x2 k xl 0
2 2
has real roots.

1-3) Solve the equations,
2 x1 4 3
a)   , b) x1 1
x 3 3 x
(Ans: a) r 2, r 3, b) r 1i 2)
1 2 1,2

1-4) Solve the equation x2 3x 42x 0
1 17
(Ans: x1 or x  )
2