Math 135 Circles and Completing the Square Examples



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

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

Algebra Review. How well do you remember your algebra?

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

Factoring Polynomials

MATH 150 HOMEWORK 4 SOLUTIONS

Operations with Polynomials

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

SPECIAL PRODUCTS AND FACTORIZATION

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

Section 7-4 Translation of Axes

Graphs on Logarithmic and Semilogarithmic Paper

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

4.11 Inner Product Spaces

Reasoning to Solve Equations and Inequalities

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

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

Vectors Recap of vectors

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

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

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

Integration by Substitution

19. The Fermat-Euler Prime Number Theorem

Repeated multiplication is represented using exponential notation, for example:

Lecture 3 Gaussian Probability Distribution

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

Binary Representation of Numbers Autar Kaw

Econ 4721 Money and Banking Problem Set 2 Answer Key

6.2 Volumes of Revolution: The Disk Method

Warm-up for Differential Calculus

Unit 6: Exponents and Radicals

Basic Analysis of Autarky and Free Trade Models

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

3 The Utility Maximization Problem

EQUATIONS OF LINES AND PLANES

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

Experiment 6: Friction

5.6 POSITIVE INTEGRAL EXPONENTS

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

Homework 3 Solutions

Review guide for the final exam in Math 233

AREA OF A SURFACE OF REVOLUTION

Integration. 148 Chapter 7 Integration

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

Small Business Networking

Regular Sets and Expressions

1.00/1.001 Introduction to Computers and Engineering Problem Solving Fall Final Exam

MODULE 3. 0, y = 0 for all y

Small Business Networking

CHAPTER 11 Numerical Differentiation and Integration

Lectures 8 and 9 1 Rectangular waveguides

Redistributing the Gains from Trade through Non-linear. Lump-sum Transfers

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

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

Helicopter Theme and Variations

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

Section 5-4 Trigonometric Functions

Understanding Basic Analog Ideal Op Amps

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

The Velocity Factor of an Insulated Two-Wire Transmission Line

1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?

Applications to Physics and Engineering

g(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany

Physics 43 Homework Set 9 Chapter 40 Key

COMPONENTS: COMBINED LOADING

MULTIPLYING OUT & FACTORING

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

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

Small Business Networking

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

Lecture 5. Inner Product

. 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

Small Business Networking

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

How To Network A Smll Business

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

Rate and Activation Energy of the Iodination of Acetone

All pay auctions with certain and uncertain prizes a comment

Rotating DC Motors Part II

PHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS

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

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006

Novel Methods of Generating Self-Invertible Matrix for Hill Cipher Algorithm

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

Review Problems for the Final of Math 121, Fall 2014

1.2 The Integers and Rational Numbers

QUADRATURE METHODS. July 19, Kenneth L. Judd. Hoover Institution

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

Data replication in mobile computing

0.1 Basic Set Theory and Interval Notation

NQF Level: 2 US No: 7480

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

Basically, logarithmic transformations ask, a number, to what power equals another number?

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

AA1H Calculus Notes Math1115, Honours John Hutchinson

Week 7 - Perfect Competition and Monopoly

ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS

Transcription:

Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for finding b when is n expression. For instnce, you should remember tht 2 + b + b 2 is perfect squre, becuse it is exctly + b) 2. How would you turn the expression x 2 + x into perfect squre? A moment of thought should convince you tht if we dd 2 )2 to x 2 + x we obtin perfect squre, becuse x + 2 )2 = x 2 + x + 2 )2. The ddition of 2 )2 is clled completing the squre, becuse the new expression cn now be written s squre of some other expression. Exmple 1. Complete the squre: x 2 + 4x = 0 x 2 + 4x = 0 x 2 + 4x + 4) = 4 x + 2) 2 = 4 We hve dded the squre of hlf the coefficient of x to the originl eqution, nd therefore to mintin equlity it ws necessry to dd the sme mount to the other side of the eqution. Wrning 2. The coefficient of x 2 must be equl to 1 in order to complete the squre. Exmple 3. Complete the squre: 2x 2 + 8x = 0 2x 2 + 8x = 0 2x 2 + 4x) = 0 2x 2 + 4x + 4) = 8 2x + 2) 2 = 8 We dded 4, the squre of hlf the coefficient of x, inside the prentheses. Note tht this mounts to dding 8 to the left side of the eqution, becuse everything inside the prentheses is multiplied by 2. Therefore, to mintin equlity we dd 8 to the right side of the eqution. In cse we cnnot set our expression equl to 0, we must subtrct whtever number we dd to the expression: Exmple 4. Complete the squre: 2x 2 + 8x 2x 2 + 8x = 2x 2 + 4x) = 2x 2 + 4x + 4) 8 = 2x + 2) 2 8 Exmple 5. x h) 2 + y k) 2 = r 2 is the eqution of circle of rdius r centered t the point h, k). Using the method of completing the squre twice) find the rdius nd center of the circle given by the eqution x 2 + y 2 + 8x 6y + 21 = 0. x 2 + y 2 + 8x 6y + 21 = 0 1) x 2 + 8x) + y 2 6y) = 21 2) x 2 + 8x + 16) + y 2 6y + 9) = 21 + 16 + 9 3) x + 4) 2 + y 3) 2 = 4 4) We hve now the form x 4)) 2 + y 3) 2 = 2 2 which is circle of rdius r = 2 centered t the point h, k) = 4, 3). University of Hwi i t Māno 52 R Spring - 2014

Mth 135 Circles nd Completing the Squre Exmples Deriving the Qudrtic Formul Given qudrtic eqution, i.e. n eqution of this form: x 2 + bx + c = 0, 0 5) where, b, nd c re rel numbers, we wish to hve formul tht will give us the explicit vlues of x for which the qudrtic eqution is zero. Tht is, we need formul tht produces x 1 nd x 2 such tht x 2 1 + bx 2 + c = 0 nd x 2 2 + bx 2 + c = 0 6) The qudrtic formul tells us exctly how to find our set of solutions {x 1, x 2 }, but it lso tells how lrge this set is. We cn hve two distinct solutions nd this hppens whenever the discriminnt is positive number. We cn hve just one solution if the discriminnt is zero. In this cse we sy tht the root x 1 = x 2 ) hs multiplicity 2, becuse it occurs twice. Finlly, when the discriminnt is negtive number, we hve squre root of negtive number nd hence no rel) solutions. Recll the qudrtic formul: x = b ± b 2 4c where the discriminnt is equl to b 2 4c 7) How do we know tht this is indeed correct? We cn pply the method of completing the squre to our qudrtic eqution 1) nd verify tht eqution 2) is correct. Here re the detils: x 2 + bx + c = 0 8) x 2 + b x + c = 0 9) x 2 + b x = c 10) ) x 2 + b 2 ) 2 b b x + = c 2 2 x 2 + b ) 2 ) 2 b b x + = c ) 2 4 c 2 ) 2 c 4 4 11) 12) 13) 14) ) 2 4 4c 15) 2 ) 2 = 16) = ± 17) = ± 18) 4 2 = ± 19) University of Hwi i t Māno 53 R Spring - 2014

Mth 135 Circles nd Completing the Squre Exmples x = b + ± x = b ± b 2 4c So which of the bove steps do we cll completing the squre? The nswer is 4) to 7); the rest del with writing the eqution in the form x = something. Let s review: Suppose you re given your fvorite qudrtic x 2 + bx + c nd need to solve for x. You re no longer mused by fctoring nd decide to complete the squre insted. Step 1: Check the coefficients. If = 0 you don t need to complete the squre. If 1 then you need to fctor out. So suppose tht 1 nd 0. [ x 2 + bx + c = x 2 + b x + c ] 22) Step 2: Group the x terms together. You complete the squre only on the terms contining the vrible x. Notice tht inside the brckets [ ] we now hve new qudrtic eqution with coefficients = 1, b = b nd c = c. x 2 + bx + c = [x 2 + b ) x + c ] Step 3: Complete the squre: dd the squre of hlf of the coefficient of x to the terms in side the prentheses ). [ x 2 + bx + c = x 2 + b ) ) ] 2 b x + + c 24) Step 4: Up until now we hve not ltered the eqution, but dding something to the right side requires subtrcting the sme number. We hve dded b2 inside the brckets [ ] nd everything inside [ ] is multiplies by. Therefore, to keep the eqution unchnged, we now subtrct from the right side the number b2 nd obtin [ x 2 + bx + c = x 2 + b ) ) ] 2 b x + + c b2 25) 4 20) 21) 23) Step 5: Simplify. The term in the prentheses ) is perfect squre nd so [ x 2 + bx + c = ) ] 2 + c b2 4 = ) 2 4 26) This form should look fmilir. If we were to set line 22) equl to zero we would hve the stndrd qudrtic eqution. Then dividing by legl since 0) nd moving terms round returns us to eqution 12). University of Hwi i t Māno 54 R Spring - 2014

Mth 135 Circles nd Completing the Squre Exmples Viete s Equtions, or how to pick out the correct pir of solutions to qudrtic eqution... Proposition 6. Given qudrtic eqution with rel coefficients, b, c x 2 + bx + c = 0, 0 If the solutions exist, then they hve the following form x 1 = b + b 2 4c x 2 = b b 2 4c nd they obey the following lgebric equtions: x 1 + x 2 = b x 1 x 2 = c If you re given qudrtic eqution to solve nd re llowed to use the qudrtic formul, then you my follow these steps nd sve yourself some work. Step 1: Mke sure tht the solutions exist, i.e. b 2 4c 0 Step 2: Look t the qudrtic eqution you hve to solve nd determine the vlues of, b, c nd compute b nd c. Step 3: Compute x 1 +x 2 nd x 1 x 2 for ech set of solutions your re given s choice. Step 4: Compre the results of steps 2 nd 3. If you find mtch, you hve found the solution. If there is no mtch, then none of the possible choices is solution. Is it possible to hve more thn one set of mtching solutions?) University of Hwi i t Māno 55 R Spring - 2014

Mth 135 Circles nd Completing the Squre Worksheet Using the method of completing the squre, put ech circle into the form. x h) 2 + y k) 2 = r 2 Then determine the center nd rdius of ech circle. 1. x 2 + y 2 10x + 2y + 17 = 0. 2. x 2 + y 2 + 8x 6y + 16 = 0. 3. 9x 2 + 54x + 9y 2 18y + 64 = 0. 4. 4x 2 4x + 4y 2 59 = 0. y h, k + r) h r, k) Center t h, k) h + r, k) Rdius r x h, k r) Smple Midterm 3 A B C D 8 A B C D 11 A B C D 29 A B C D 31 A B C D 32 A B C D 35 A B C D Smple Finl University of Hwi i t Māno 56 R Spring - 2014