math cad prime quadratic equation derivation

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MathCAD Prime - Quadratic Equation Derivation.mcdx AlternateSolutionoftheQuadraticSolutionDerivation byJulioC.Banks,MSME,PE EstablishthemostpracticalformofthesolutionofQuadraticEquations + + ax 2 bx c 0 ( (1) ) DivideEq.1throuhgbythecoefficientof ,i.e.,a x 2 + + x 2 b a x c a 0 ( (2) ) Let 2 B 1 b a ( (3) ) and B 0 c a ( (4) ) Solvefor fromEq.3 B 1 B 1 1 2 b a ( (5) ) SubstituteEq.3and4intoEq.2 + + x 2 2 B 1 x B 0 0 ( (6) ) Thefirsttwo(2)termsofEq.6resemblesthebinomialform + + x 2 2 B 1 x B 1 2 + x B 1 2 ( (7) ) Therefore,transformEq.6bysubtracting frombothsidesofEq.6andadding toboth B 0 B 1 2 sidesofEq.6 + + x 2 2 B 1 x B 1 2 - B 1 2 B 0 ( (8) ) JulioC.Banks,PE Page1of2

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Page 1: Math cad prime   quadratic equation derivation

MathCAD Prime - Quadratic Equation Derivation.mcdx

Alternate Solution of the Quadratic Solution Derivationby Julio C. Banks, MSME, PE

Establish the most practical form of the solution of Quadratic Equations

=++⋅a x2

⋅b x c 0 ((1))

Divide Eq. 1 throuhg by the coefficient of , i.e., ax2

=++x2

⋅⎛⎜⎝―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

=++x2

⋅⋅2 B1 x B0 0 ((6))

The first two (2) terms of Eq. 6 resembles the binomial form

=++x2

⋅⋅2 B1 x B1

2 ⎛⎝ +x B1⎞⎠2

((7))

Therefore, transform Eq. 6 by subtracting from both sides of Eq. 6 and adding to both B0 B1

2

sides of Eq. 6

=++x2

⋅⋅2 B1 x B1

2−B1

2B0 ((8))

Julio C. Banks, PE Page 1 of 2

Page 2: Math cad prime   quadratic equation derivation

MathCAD Prime - Quadratic Equation Derivation.mcdx

Equating 7 and 8

=⎛⎝ +x B1⎞⎠2

−B1

2B0 ((9))

Solve for from Eq. 9x

=x +−B1

‾‾‾‾‾‾‾2

−B1

2B0

− ‾‾‾‾‾‾‾2

−B1

2B0

⎡⎢⎢⎣

⎤⎥⎥⎦

((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

Page 3: Math cad prime   quadratic equation derivation

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 M ike Gover, University of Bradford

INDUSTRIAL PRESS, INC. NEW YORK

Page 4: Math cad prime   quadratic equation derivation

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

Page 5: Math cad prime   quadratic equation derivation

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 of the formula: it is important

As an example, we shall solve the equation 2X2 - 3x - 4 = O.

-b± y'bC 4acHere 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 = ....... . ... .