The principal reason for this presentation is that one can often choose, a priori, the root that has physical meaning such as positive geometry entities such as positive dimensions, etc

1. MathCAD Prime - Quadratic Equation Derivation.mcdx
Alternate Solution of the Quadratic Solution Derivation
by Julio C. Banks, MSME, PE
Establish the most practical form of the solution of Quadratic Equations
＝++⋅a x
2
⋅b x c 0 ((1))
Divide Eq. 1 throuhg by the coefficient of , i.e., ax
2
＝++x
2
⋅
⎛
⎜⎝
―
b
a
⎞
⎟⎠
x
⎛
⎜⎝
―
c
a
⎞
⎟⎠
0 ((2))
Let ＝⋅2 B1 ―
b
a
((3))
and ＝B0 ―
c
a
((4))
Solve for from Eq. 3B1
＝B1 ⋅―
1
2
⎛
⎜⎝
―
b
a
⎞
⎟⎠
((5))
Substitute Eq. 3 and 4 into Eq. 2
＝++x
2
⋅⋅2 B1 x B0 0 ((6))
The first two (2) terms of Eq. 6 resembles the binomial form
＝++x
2
⋅⋅2 B1 x B1
2
⎛⎝ +x B1⎞⎠
2
((7))
Therefore, transform Eq. 6 by subtracting from both sides of Eq. 6 and adding to bothB0 B1
2
sides of Eq. 6
＝++x
2
⋅⋅2 B1 x B1
2
−B1
2
B0 ((8))
Julio C. Banks, PE Page 1 of 2

2. MathCAD Prime - Quadratic Equation Derivation.mcdx
Equating 7 and 8
＝⎛⎝ +x B1⎞⎠
2
−B1
2
B0 ((9))
Solve for from Eq. 9x
＝x +−B1
‾‾‾‾‾‾‾2
−B1
2
B0
− ‾‾‾‾‾‾‾2
−B1
2
B0
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
((10))
Equation 10 can be simplified as follows
Let ＝Λ ――
B0
B1
2
((11))
Substitute Eq. 11 into Eq. 10
＝x ⋅−B1
⎛
⎜
⎝
−1 ‾‾‾‾
2
−1 Λ
+1 ‾‾‾‾
2
−1 Λ
⎡
⎢
⎣
⎤
⎥
⎦
⎞
⎟
⎠
((12))
The author has discovered that equation 12 is most useful in determining, a priori, the single
root that would produce a physically meaningful result such as gemetric parameters, e.g.,
length, area, or volume must be positive geometric entities. That is, one is able to know in
advance which root to retain from a symbolic derivation of a problem which results in a
quadratic equation of the standard from given in equation 1.
Julio C. Banks, PE Page 2 of 2

3. EngineerIng
Mathematics
K. A. Stroud
Formerly Principal Lecturer
Department of Mathematics
Coventry University
United Kingdom
with additions by
Dexter f. Booth
Principal Lecturer
School of Computing and Mathematics
University of Huddersfield
United Kingdom
FIFTH EDITION
Review Board for the fifth edition:
Dr Charles Fall, University of Northumbria at Newcastle
Dr Pat Lewis, Staffordshire University
Dr Mark Kermode, University of Liverpool
Dr Hazel Shute, University of Plymouth
Dr Mike Gover, University of Bradford
INDUSTRIAL PRESS, INC.
NEW YORK

4. Library of Congress Cataloging-in-Publication Data
Stroud, K. A.
Engineering mathematics! KA. Stroud; with additions by Dexter J. Booth.-5th ed.
p. cm.
Includes index.
ISBN 0-8311-3152-7
l.Engineering mathematics-Programmed instruction. 1. Booth, Dexter J. II. Title.
TA330 .S78 2001
51O'.246-dc21 2001039100
Published in North America under license from Palgrave Publishers Ltd, Houndrnills, Basingstoke,
Rants RG2l 6XS, United Kingdom.
Industrial Press, Inc.
200 Madison Avenue
New York, NY 10016-4078
Copyright © 2001 by Industrial Press Inc., New York. Printed in the
United States of America. All right reserved. This book, or any parts thereof, may
not be reproduced, stored in a retrieval system, or transmitted in any form without
the permission of the publisher.
]0 9 8 7 6 5 4 3 2

5. 204 Foundation topics
3 Solution by formula
We can establish a formula for the solution of the general quadratic equation
ax2+ bx + c = 0 which is based on the method of completing the square:
ax2+ bx + c = 0
Dividing throughout by the coefficient of x, i.e. a:
2 b c
x +-x+-=O
a a
Subtracting ~ from each side gives x2 +~x = - ~a a a
We then <:ldd to each side the square of half the coefficient of x:
X2+~X+ (~)2=_~+ (~)2a 2a a 2a
b b2 b2 c
x2+-x+-=--­
a 4a2 4a2 a
(
~) 2 _ b2 _ 4ac
x + 2a - 4a2
~ _ !b2 - 4ac _ ;/b2 - 4ac b ;/b2 - 4ac
x =-- ±-~--x + 2a - ± 4a2 - ± 2a 2a 2a
-b ± ;/~b2----4-:-a-c
Ifax2+ bx + c = 0, x = 2a
Substituting the values of a, band c for any particular quadratic equation gives
the solutions of the equation.
Make a note ofthe formula: it is important
As an example, we shall solve the equation 2X2 - 3x - 4 = O.
-b± y'bC 4ac
Here a = 2 b = -3 c = -4 and x = ---=---­
" 2a
3 ± V9 - 4 x 2 x (-4) 3 ± ;/9 + 32 3 ± V4i
X= 4 = 4 = --4-­
3 ± 6·403 -3·403 9·403
4 4 or-­4
x = -0·851 or x = 2·351
It is just a case of careful substitution. You need, of course, to remember the
formula. For
ax2+bx +c +0 x = ....... .... .