Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:



Similar documents
Reasoning to Solve Equations and Inequalities

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn

SPECIAL PRODUCTS AND FACTORIZATION

MA Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!

Multiplication and Division - Left to Right. Addition and Subtraction - Left to Right.

Algebra Review. How well do you remember your algebra?

Math 135 Circles and Completing the Square Examples

EQUATIONS OF LINES AND PLANES

Pure C4. Revision Notes

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Operations with Polynomials

Factoring Polynomials

Unit 6: Exponents and Radicals

1.2 The Integers and Rational Numbers

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

MATH 150 HOMEWORK 4 SOLUTIONS

Section 7-4 Translation of Axes

CS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001

Section 5-4 Trigonometric Functions

. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator

Integration. 148 Chapter 7 Integration

Homework 3 Solutions

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

Binary Representation of Numbers Autar Kaw

9 CONTINUOUS DISTRIBUTIONS

Repeated multiplication is represented using exponential notation, for example:

Graphs on Logarithmic and Semilogarithmic Paper

Math 314, Homework Assignment Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.

Regular Sets and Expressions

Pentominoes. Pentominoes. Bruce Baguley Cascade Math Systems, LLC. The pentominoes are a simple-looking set of objects through which some powerful

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.

All pay auctions with certain and uncertain prizes a comment

Or more simply put, when adding or subtracting quantities, their uncertainties add.

AREA OF A SURFACE OF REVOLUTION

Integration by Substitution

0.1 Basic Set Theory and Interval Notation

Small Business Networking

The remaining two sides of the right triangle are called the legs of the right triangle.

4.11 Inner Product Spaces

2 DIODE CLIPPING and CLAMPING CIRCUITS

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

One Minute To Learn Programming: Finite Automata

Treatment Spring Late Summer Fall Mean = 1.33 Mean = 4.88 Mean = 3.

Econ 4721 Money and Banking Problem Set 2 Answer Key

Chapter. Contents: A Constructing decimal numbers

Small Business Networking

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation

e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

How To Network A Smll Business

Experiment 6: Friction

Helicopter Theme and Variations

and thus, they are similar. If k = 3 then the Jordan form of both matrices is

Roots of Polynomials. Ch. 7. Roots of Polynomials. Roots of Polynomials. dy dt. a dt. y = General form:

Basic Analysis of Autarky and Free Trade Models

LECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.

5.6 POSITIVE INTEGRAL EXPONENTS

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

COMPLEX FRACTIONS. section. Simplifying Complex Fractions

Vector differentiation. Chapters 6, 7

Small Business Networking

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

A.7.1 Trigonometric interpretation of dot product A.7.2 Geometric interpretation of dot product

Small Business Networking

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Thinking out of the Box... Problem It s a richer problem than we ever imagined

CHAPTER 11 Numerical Differentiation and Integration

Vectors Recap of vectors

Lecture 3 Gaussian Probability Distribution

10.6 Applications of Quadratic Equations

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity

Maximum area of polygon

Geometry 7-1 Geometric Mean and the Pythagorean Theorem

Welch Allyn CardioPerfect Workstation Installation Guide

Exponential and Logarithmic Functions

Point Groups and Space Groups in Geometric Algebra

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Acid-Base Equilibria Acid-Base Definitions

1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.

6.2 Volumes of Revolution: The Disk Method

Small Businesses Decisions to Offer Health Insurance to Employees

Words Symbols Diagram. abcde. a + b + c + d + e

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

How To Set Up A Network For Your Business

Vectors and dyadics. Chapter 2. Summary. 2.1 Examples of scalars, vectors, and dyadics

Distributions. (corresponding to the cumulative distribution function for the discrete case).

19. The Fermat-Euler Prime Number Theorem

, and the number of electrons is -19. e e C. The negatively charged electrons move in the direction opposite to the conventional current flow.

MATH PLACEMENT REVIEW GUIDE

Solving BAMO Problems

AntiSpyware Enterprise Module 8.5

COMPONENTS: COMBINED LOADING

Quick Reference Guide: One-time Account Update

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

Lecture 5. Inner Product

Transcription:

Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you to find the solutions of equtions. However, qudrtics re not lwys esy to fctor. Sometimes qudrtics cnnot e fctored completely. Two procedures tht cn help you to fctor qudrtic re completing the squre nd the qudrtic formul. Completing the squre In Appendi A, two specil cses of epnding rckets were considered: ( + ) = + + ( ) = +. These were clled perfect squres. In Appendi C, two strtegies suggested for fctoring lgeric epressions were to look for perfect squres nd to look for differences of squres. Emple D.1 Fctor the qudrtic epression: + 6 + 8 y looking for perfect squres nd differences of squres. Solution: The given lgeric epression is very close to the perfect squre: ( + 3) = + 6 + 9. So: + 6 + 8 = ( + 6 + 9) 1 = ( + 3) 1. Since 1 = 1, the lst epression is difference of two squres, nd so: + 6 + 8 = ( + 3) 1 = ( + 3 +1)( + 3 1) = ( + 4)( ). Emple D.1 is very elorte nd counter-intuitive wy to fctor stright-forwrd lgeric epression like + 6 + 8. The process outlined in Emple D.1 hs one gret dvntge over quicker, more intuitive methods: the method of Emple D.1 will work

when the numers involved re more complicted nd it is hrder to fctor the qudrtic epression. Emple D. Fctor the qudrtic epression: + 66 + 8 y looking for perfect squres nd differences of squres. Solution: The perfect squre tht this qudrtic epression is relted to is: ( + 33) = + 66 + 1089. The qudrtic epression tht we hve to fctor is relted to this perfect squre: + 66 + 8 = ( + 66 +1089) 1081 = ( + 33) 1081. This lst lgeric epression is difference of two squres, so: + 66 + 8 = ( + 33 + 3.879)( + 33 3.879) = ( + 65.879)( + 0.11). The numer 3.879 ppers ecuse the squre root of 1081 is pproimtely 3.879. The process of fctoring qudrtic equtions illustrted in Emples D.1 nd D. is clled completing the squre. As indicted, not ll qudrtics cn e completely fctored. The process of completing the squre cn tell you when this is the cse. Emple D.3 Try to fctor the qudrtic epression: + 6 + 10 y looking for perfect squres nd differences of squres. Solution: Following the pttern set out in Emples D.1 nd D., you could strt y finding perfect squre tht is closely relted to + 6 + 10. The perfect squre: ( + 3) = + 6 + 9 is very closely relted to + 6 + 10. In fct, + 6 + 10 = ( + 6 + 9) + 1 = ( + 3) + 1.

This is where the pttern estlished in Emples D.1 nd D. reks down, ecuse the epression tht we hve is not difference of two squres. The epression: ( + 3) + 1 cnnot e fctored ny further ecuse it is the sum rther thn the difference of two squres. The Qudrtic Formul The process of fctoring qudrtic epression y completing the squre cn e summrized s n lgeric formul. The formul is this: If you re trying to fctor the qudrtic epression: + + c, then the fctors re: c c + + + 4 4. There is one importnt cvet when using this method to fctor polynomil: ecuse you hve to tke squre root, the quntity: 4c must e greter thn or equl to zero. If 4c is negtive, then it is not possile to fctor the qudrtic epression + + c. Emple D.4 Use the qudrtic formul to fctor the following qudrtic epressions. If it is not possile to fctor ny of the qudrtic epressions, indicte why you think this to e the cse. ) 6 + 9 +. ) + + 1. c) + + 1.

Solution: ) To fctor 6 + 9 +, note the similrities etween this epression nd + + c. The correspondence is tht: = 6. = 9. c =. Before plugging into the formul, it is wise to mke sure tht the quntity 4c is greter thn or equl to zero. (If the quntity 4c is negtive, then the qudrtic epression cnnot e fctored, so there will e no sense in trying.) 4c = 9 4 6 = 81 48 = 33. The numer 33 is greter thn or equl to zero, so the qudrtic will fctor. Plugging = 6, = 9 nd c = into the formul gives: 6 + 9 + = 6( + 1.87)( +0.713). ) To fctor + + 1, gin note the similrities etween this epression nd + + c. In this cse, the correspondence is tht: = 1. = 1. c = 1. To check whether or not the epression + + 1, cn e fctored, you cn check the sign of 4c. In this cse, 4c = 1 4 1 1 = 1 4 = 3. Since 4c is negtive, the qudrtic epression + + 1 cnnot e fctored. c) Here the nlysis is just like the previous two cses, ecept tht: = 1. =. c = 1. Checking the quntity 4c gives 4c = 4 4 1 1 = 0. As this is not negtive, the qudrtic epression + + 1 will fctor. Plugging = 1, = nd c = 1 into the formul gives: ( + 1)( + 1) = ( + 1). An lterntive (nd eqully vlid) wy to fctor this prticulr qudrtic epression would hve een to relize tht + + 1 ws perfect squre, so tht you cn determine tht + + 1 = ( + 1) without hving to use the formul.

Using Completing the Squre to Otin the Qudrtic Formul You might wonder how the formul for the fctors of the qudrtic epression + + c ws otined. The working shown elow indictes how the process of completing the squre my e used to otin the formul. The working presented elow is quite formidle ecuse it fetures lot of symols, rther thn just concrete numers. Use the eplntory notes for ech step to follow wht is going on. + + c = c + + = c + + + = c + Fctor out the. Mke the epression s much like the perfect squre: + possile. s Fctor the perfect squre nd group the left-over terms together. c = + Simplify the term: 4 =. 4 = + 4c Put the two terms over common 4 denomintor. (See Appendi E.) Relize tht wht you hve is difference of two squres. = + + 4c + 4 4c Use the difference of squres to 4 fctor. = + + 4c + 4c Tke the squre roots of the numertors nd denomintors. c c = + + + 4 4. Comine like terms.

Eercises for Appendi D For Prolems 1-10, complete the squre for the given qudrtic epressions. 1. y + y.. + 8 + 5. 3. u 14u. 4. 4 + 4 + 1. 5. 3r + 1r + 16. 6. +10 +1. 7. t 7t 8. 8.. 9. 3r 4 r + 14. 10. w + 3w. For Prolems 11-15, fctor the qudrtic epressions (if possile). 11. r + 7r + 1. 1. y + 5y +3. 13. e + 1 + 4e + 1 + 4e. 14. 3t + 13t + 9. 15. u + 4u 5.

Answers to Eercises for Appendi D 1. (y + 1) - 1.. ( + ) - 3. 3. (u - 7) - 49. 4. ( + ) - 3. 5. 3(r + ) + 4. 6. -( - 5) + 4. 7. (t - 3.5t) - 0.5. 8.. 9. 3(r - 4) - 34. 10. (w + 3/) -.5. 11. (r + 3)(r + 4). 1. (y + 1)(y + 3/). 13. e(e + ). 14. 3(t + 3.468)(t + 0.865). 15. This qudrtic does not fctor. To see this, - 4c = 16-4(-1)(-5) = -4.