DOCSLIB.ORG

Quadratic Equations and Conics

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Quadratic Equations and Conics Quadratic Equations and Conics A quadratic equation in two variables is an equation that’s equivalent to an equation of the form p(x, y)=0 where p(x, y)isaquadraticpolynomial. Examples. 4x2 3xy 2y2 + x y +6=0isaquadraticequation,asare •x2 y−2 =0and− x2 + y−2 =0andx2 1=0. − − y = x2 is a quadratic equation. It’s equivalent to y x2 =0, •and y x2 is a quadratic polynomial. − − xy =1isaquadraticequation.It’sequivalenttothequadratic •equation xy 1=0. − x2 +y2 = 1isaquadraticequation.It’sequivalenttox2 +y2 +1 = 0. • − x2 + y = x2 +2isanotaquadraticequation.It’salinearequation. •It’s equivalent to y 2=0,andy 2isalinearpolynomial. − − A conic is a set of solutions of a quadratic equation in two variables. In contrast to lines–solutions of linear equations in two variables–it takes a fair amount of work to list all of the possible geometric shapes that can possibly arise as conics. In what remains of this chapter, we’ll take a tour of some conics that we already know. After we cover trigonometry in this course, we’ll return to conics and explain all of the possible shapes of conics. 154 Parabola The graph of the function in one variable f(x) = x2 is called a parabola. No points We saw in the chapter PolynomialJ Equations that the quadratic equation x2 + y2 = 1 − has no solution. A sum of squares can never be negative. Thus, is an example of conic. ; x2 = 1hasnosolutiontoo! − The poilits iii the graphOneof pointf(x) are points of the form (x, f(x)). That is, the poillts in the graph haveWe sawthe inform the chapter(x, x2). PolynomialThese are the Equationspoints in thatthe the quadratic equation plane whose second coordinates are the square of theirxfirst2 + ycoordinates,2 =0 or in other words, points of the form (x, y) where has exactly one solution, the point (0, 0). The only way to take two numbers, square them,y take= the sum of those squares, and get 0, is if the two numbers To recap, the graph youof f(x) start= withx2 in arethe 0.plane Thus,is the thesame set containingset as the theset single point (0, 0) is a of solutions of the quadraticconic.equation in two variables given by the equation = x2 0 I 150 155 ParallelParallellines lines AnAnequation equationfor forthe thevertical verticalline linethat thatcrosses crossesthe thex-axisx-axisat atthe thenumber number1 1 isis x==1,orequivalently,1, equivalently, xx— 11 = 0.Similarly, Similarly,xx +1=0isanequationfor1 = 0 is an equation for x or − 0. + thethevertical verticalline linethat thatcrosses crossesthe thex-axisx-axisat atthe thenumber number—1.1.As Aswe wesaw sawin inthe the − previouspreviouschapter, chapter,the theunion unionof ofthese thesetwo twovertical verticallines linesis isthe theset setof ofsolutions solutions ofofthe theequation equation (x— 1)(x1)(x +1)=01) = (x − + 0 ananequation equationthat thatcan canbe bewritten writtenmore moresimply simplyas as x21—Ox2 1=0 − Thus,Thus,two twoparallel parallellines linescami canappear appearas asa aconic. conic. —1 Parallel lines IntersectingIntersectingAn equation forlines linesthe vertical line that crosses t.he x-axis at the number 1 = — = isIt’sIt’sx also also1, possibleor possibleequivalently,to tohave havextwo two1intersecting intersecting0. Similarly,lines linesxas as+a a1 conic. conic.0 is In Inanthe theequationprevious previousfor chapterchapterthe verticalwe wesaw sawhuethat thatthatthe thecrossessolutions solutionsthe x-axisof of at the number —1. As we saw in the previous chapter, the union of these two vertical lines is the set of solutions x2— y2 ==0 of the equation − 0 are a union of the two lines that have slope 1 and 1, and that each pass are a union of the two lines that(x — have1)(x +slope1) =1 0and —1, and that each pass through the point (0, 0). Recall that x y =0isanequationforthefirstof− throughan equationthe pointthat can(0, 0).be Recallwrittenthatmorex —simplyy = 0 asis an equation for the first of these two lines, and that x + y==0isanequationforthesecondofthesetwo− these two lines, and that x + y 0 is an equation= for the second of these two lines. Therefore, (x y)(x + y)= = 0 is—1 an equation0 for the union of these two lines. Therefore, (x —−y) (x + y) 0 is an equation for the union of these two lines,Thus, andtwo itparallel can belines simplifiedcan appear as x2 asya2 ==0.conic. lines, and it can be simplified as —− 0. / N 156 Intersecting lines 149 It’s also possible to have two intersecting lines as a conic. In the previous chapter we saw that the solutions of 2 —y_=O are a union of the two lines that have slope 1 and —1, and that each pass through the point (0, 0). Recall that x — y 0 is an equation for the first of these two hues, and that x + y = 0 is an equation for the second of these two lines. Therefore, (x — y)(x + y) = 0 is an equation for the union of these two lines, and it can be simplified as x2 — y2 = 0. 149 Single line x2 =0isaquadraticequation,soitssetofsolutionsintheplaneisaconic. The only number whose square is 0 is 0 itself, so if x2 =0,thenx =0. Thus, the set of solutions of x2 = 0 are those points in the plane whose x- coordinates equal 0. This is a single vertical line, the y-axis. This single line is an example of a conic. _____ __ ___ Parabola The graph ofParabolathe function in one variable f(x) = x2 is called a parabola. The graph of the function in one variable f(x)=x2 is called a parabola. J(x=x2 The points in the graph of f(x)arepointsoftheform(x, f(x)). That is, The points inthethe pointsgraph inof thef(x) graphare points have theof the formform (x,(x, x2).f(x)). TheseThat areis, the points in the the points in theplanegraph whosehave secondthe form coordinates(x, x2). areThese theare squarethe ofpoints theirin firstthe coordinates, or plane whose secondin othercoordinates words, pointsare the of thesquare formof (x,their y)wherefirst coordinates, or in other words, points of the form (x, y) where y = x2 = 157 = To recap, the graph of f(x) x2 in the plane is the same set as the set of solutions of the quadratic equation in two variables given by the equation = J 150 _______ _______ _______ _______ _______ _______ Parabola The graph of the function in one variable f(x) = x2 is called a parabola. To recap, the graph of f(x)=x2 in the plane is the same set as the set Parabola of solutions of the quadratic equation in two variables given by the equation 2 The graph ofythe= xfunction. That is,in theone parabolavariable f(x) below= isx2 ais conic.called a parabola. The points in the graph of f(x) are points of the form (x, f(x)). That is, the points in the graph have the form (x, x2). These are the points in the plane whose second coordinates are the square of their first coordinates, or in other words, points of the form (x, y) Iiere The points in the graph of f(x) are points of the form (x, f(x)). That is,= the points in the graph have the form x2). These are Hyperbola (x, the points in= the plane whose second coordinates are the squareTo recap,of theirthe graphfirst of f(x) x2orin the plane is the same set as the set coordinates,1 1 in other words, pointsThe pointsof the inform the graph(x, y)ofIiere ofsolutions the functionof thegquadratic(x)=x areequation of the formin two (x,variablesx). given by the equation = y=12. = To recap, the graph of f(x) x2 in the plane is the same set as the set .of solutions of the quadratic equation in two variables given by the equation y=12 JA(2)(s) JA(2)(s) a a That is, they are points whose y-coordinates are the multiplicative150 inverse of 1 their x-coordinates. In other words, the graph of g(x)=x is exactly the set 1 of points (x, y)intheplanewherey = x. 158 150 _____ __ ___ Parabola The graph of the function in one variable f(x) = x2 is called a parabola. J(x=x2 The points in the graph of f(x) are points of the form (x, f(x)). That is, the points in the graph have the form (x, x2). These are the points in the 1 We know from the domainplane ofwhose the functionsecond coordinatesg(x)=x thatare xthe=0.Thus,wesquare of their first coordinates, or 1 6 can multiply the equationin othery = xwords,by x topoints obtainof thethe equivalentform (x, y) equationwhere xy =1 = 1 = x2 so that we now have thatTo therecap, graphthe of xgraphis exactlyof f(x) the set ofin solutionsthe plane of theis the same set as the set equation xy =1. of solutions of the quadratic equation in two variables given by the equation = J The equation xy =1isaquadraticequation,soitssetofsolutionsisa150 conic. It’s called a hyperbola. 159 _______ _______ _______ Exercises ParabolaParabola #1-5 involve writing equations for planar transformations of the parabola that is the set of solutions of y = x2. We’ll call this conic S. The graph of the function in one variable f(x) = x2 is called The graph of the function in one variable f(x) = x2 is calledJ2a parabola.a parabola. The points in the graph of are points of the form The points in the graph1.)ofThef(x) parabolaaref()points shiftedof the form right(x, 2f(x)). and(x, upThatf(x)). 4: is, That is, the points in the graph have x2). 2 2 the form (x, These are the points in the the points in the graphRecallhave thatthe form the planar(x, x2).
Load more

Recommended publications
  • The Diamond Method of Factoring a Quadratic Equation
    The Diamond Method of Factoring a Quadratic Equation Important: Remember that the first step in any factoring is to look at each term and factor out the greatest common factor. For example: 3x2 + 6x + 12 = 3(x2 + 2x + 4) AND 5x2 + 10x = 5x(x + 2) If the leading coefficient is negative, always factor out the negative. For example: -2x2 - x + 1 = -1(2x2 + x - 1) = -(2x2 + x - 1) Using the Diamond Method: Example 1 2 Factor 2x + 11x + 15 using the Diamond Method. +30 Step 1: Multiply the coefficient of the x2 term (+2) and the constant (+15) and place this product (+30) in the top quarter of a large “X.” Step 2: Place the coefficient of the middle term in the bottom quarter of the +11 “X.” (+11) Step 3: List all factors of the number in the top quarter of the “X.” +30 (+1)(+30) (-1)(-30) (+2)(+15) (-2)(-15) (+3)(+10) (-3)(-10) (+5)(+6) (-5)(-6) +30 Step 4: Identify the two factors whose sum gives the number in the bottom quarter of the “x.” (5 ∙ 6 = 30 and 5 + 6 = 11) and place these factors in +5 +6 the left and right quarters of the “X” (order is not important). +11 Step 5: Break the middle term of the original trinomial into the sum of two terms formed using the right and left quarters of the “X.” That is, write the first term of the original equation, 2x2 , then write 11x as + 5x + 6x (the num bers from the “X”), and finally write the last term of the original equation, +15 , to get the following 4-term polynomial: 2x2 + 11x + 15 = 2x2 + 5x + 6x + 15 Step 6: Factor by Grouping: Group the first two terms together and the last two terms together.
    [Show full text]
  • Solving Cubic Polynomials
    Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial monic. a 2. Then, given x2 + a x + a , substitute x = y − 1 to obtain an equation without the linear term. 1 0 2 (This is the \depressed" equation.) 3. Solve then for y as a square root. (Remember to use both signs of the square root.) a 4. Once this is done, recover x using the fact that x = y − 1 . 2 For example, let's solve 2x2 + 7x − 15 = 0: First, we divide both sides by 2 to create an equation with leading term equal to one: 7 15 x2 + x − = 0: 2 2 a 7 Then replace x by x = y − 1 = y − to obtain: 2 4 169 y2 = 16 Solve for y: 13 13 y = or − 4 4 Then, solving back for x, we have 3 x = or − 5: 2 This method is equivalent to \completing the square" and is the steps taken in developing the much- memorized quadratic formula. For example, if the original equation is our \high school quadratic" ax2 + bx + c = 0 then the first step creates the equation b c x2 + x + = 0: a a b We then write x = y − and obtain, after simplifying, 2a b2 − 4ac y2 − = 0 4a2 so that p b2 − 4ac y = ± 2a and so p b b2 − 4ac x = − ± : 2a 2a 1 The solutions to this quadratic depend heavily on the value of b2 − 4ac.
    [Show full text]
  • Algebra II Level 1
    Algebra II 5 credits – Level I Grades: 10-11, Level: I Prerequisite: Minimum grade of 70 in Geometry Level I (or a minimum grade of 90 in Geometry Topics) Units of study include: equations and inequalities, linear equations and functions, systems of linear equations and inequalities, matrices and determinants, quadratic functions, polynomials and polynomial functions, powers, roots, and radicals, rational equations and functions, sequence and series, and probability and statistics. PROFICIENCIES INEQUALITIES -Solve inequalities -Solve combined inequalities -Use inequality models to solve problems -Solve absolute values in open sentences -Solve absolute value sentences graphically LINEAR EQUATION FUNCTIONS -Solve open sentences in two variables -Graph linear equations in two variables -Find the slope of the line -Write the equation of a line -Solve systems of linear equations in two variables -Apply systems of equations to solve real word problems -Solve linear inequalities in two variables -Find values of functions and graphs -Find equations of linear functions and apply properties of linear functions -Graph relations and determine when relations are functions PRODUCTS AND FACTORS OF POLYNOMIALS -Simplify, add and subtract polynomials -Use laws of exponents to multiply a polynomial by a monomial -Calculate the product of two or more polynomials -Demonstrate factoring -Solve polynomial equations and inequalities -Apply polynomial equations to solve real world problems RATIONAL EQUATIONS -Simplify quotients using laws of exponents -Simplify
    [Show full text]
  • Math: Honors Geometry UNIT/Weeks (Not Timeline/Topics Essential Questions Consecutive)
    Math: Honors Geometry UNIT/Weeks (not Timeline/Topics Essential Questions consecutive) Reasoning and Proof How can you make a Patterns and Inductive Reasoning conjecture and prove that it is Conditional Statements 2 true? Biconditionals Deductive Reasoning Reasoning in Algebra and Geometry Proving Angles Congruent Congruent Triangles How do you identify Congruent Figures corresponding parts of Triangle Congruence by SSS and SAS congruent triangles? Triangle Congruence by ASA and AAS How do you show that two 3 Using Corresponding Parts of Congruent triangles are congruent? Triangles How can you tell whether a Isosceles and Equilateral Triangles triangle is isosceles or Congruence in Right Triangles equilateral? Congruence in Overlapping Triangles Relationships Within Triangles Mid segments of Triangles How do you use coordinate Perpendicular and Angle Bisectors geometry to find relationships Bisectors in Triangles within triangles? 3 Medians and Altitudes How do you solve problems Indirect Proof that involve measurements of Inequalities in One Triangle triangles? Inequalities in Two Triangles How do you write indirect proofs? Polygons and Quadrilaterals How can you find the sum of The Polygon Angle-Sum Theorems the measures of polygon Properties of Parallelograms angles? Proving that a Quadrilateral is a Parallelogram How can you classify 3.2 Properties of Rhombuses, Rectangles and quadrilaterals? Squares How can you use coordinate Conditions for Rhombuses, Rectangles and geometry to prove general Squares
    [Show full text]
  • Quadratic Equations Through History
    Cal McKeever & Robert Bettinger COMMON CORE STANDARD 3B Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. Factor a quadratic expression to reveal the zeros of the function it defines. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. SYNOPSIS At a summit of time-traveling historical mathematicians, attendees from Ancient Babylon pose the question of how to solve quadratic equations. Their method of completing the square serves their purposes, but there was far more to be done with this complex equation. Pythagoras, Euclid, Brahmagupta, Bhaskara II, Al-Khwarizmi and Descartes join the discussion, each pointing out their contributions to the contemporary understanding of solving quadratics through the quadratic formula. Through their discussion, the historical situation through which we understand the solving of quadratic equations is highlighted, showing the complex history of this formula taught everywhere. As the target audience for this project is high school students, the different characters of the script use language and symbolism that is anachronistic but will help the the students to understand the concepts that are discussed. For example, Diophantus did not use a,b,c in his actual texts but his fictional character in the the given script does explain his method of solving quadratics using contemporary notation. HISTORICAL BACKGROUND The problem of solving quadratic equations dates back to Babylonia in the 2nd Millennium BC. The Babylonian understanding of quadratics was used geometrically, to solve questions of area with real-world solutions.
    [Show full text]
  • The Computational Complexity of Polynomial Factorization
    Open Questions From AIM Workshop: The Computational Complexity of Polynomial Factorization 1 Dense Polynomials in Fq[x] 1.1 Open Questions 1. Is there a deterministic algorithm that is polynomial time in n and log(q)? State of the art algorithm should be seen on May 16th as a talk. 2. How quickly can we factor x2−a without hypotheses (Qi Cheng's question) 1 State of the art algorithm Burgess O(p 2e ) 1.2 Open Question What is the exponent of the probabilistic complexity (for q = 2)? 1.3 Open Question Given a set of polynomials with very low degree (2 or 3), decide if all of them factor into linear factors. Is it faster to do this by factoring their product or each one individually? (Tanja Lange's question) 2 Sparse Polynomials in Fq[x] 2.1 Open Question α β Decide whether a trinomial x + ax + b with 0 < β < α ≤ q − 1, a; b 2 Fq has a root in Fq, in polynomial time (in log(q)). (Erich Kaltofen's Question) 2.2 Open Questions 1. Finding better certificates for irreducibility (a) Are there certificates for irreducibility that one can verify faster than the existing irreducibility tests? (Victor Miller's Question) (b) Can you exhibit families of sparse polynomials for which you can find shorter certificates? 2. Find a polynomial-time algorithm in log(q) , where q = pn, that solves n−1 pi+pj n−1 pi Pi;j=0 ai;j x +Pi=0 bix +c in Fq (or shows no solutions) and works 1 for poly of the inputs, and never lies.
    [Show full text]
  • Section P.2 Solving Equations Important Vocabulary
    Section P.2 Solving Equations 5 Course Number Section P.2 Solving Equations Instructor Objective: In this lesson you learned how to solve linear and nonlinear equations. Date Important Vocabulary Define each term or concept. Equation A statement, usually involving x, that two algebraic expressions are equal. Extraneous solution A solution that does not satisfy the original equation. Quadratic equation An equation in x that can be written in the general form 2 ax + bx + c = 0 where a, b, and c are real numbers with a ¹ 0. I. Equations and Solutions of Equations (Page 12) What you should learn How to identify different To solve an equation in x means to . find all the values of x types of equations for which the solution is true. The values of x for which the equation is true are called its solutions . An identity is . an equation that is true for every real number in the domain of the variable. A conditional equation is . an equation that is true for just some (or even none) of the real numbers in the domain of the variable. II. Linear Equations in One Variable (Pages 12-14) What you should learn A linear equation in one variable x is an equation that can be How to solve linear equations in one variable written in the standard form ax + b = 0 , where a and b and equations that lead to are real numbers with a ¹ 0 . linear equations A linear equation has exactly one solution(s). To solve a conditional equation in x, . isolate x on one side of the equation by a sequence of equivalent, and usually simpler, equations, each having the same solution(s) as the original equation.
    [Show full text]
  • Quadratic Polynomials
    Quadratic Polynomials If a>0thenthegraphofax2 is obtained by starting with the graph of x2, and then stretching or shrinking vertically by a. If a<0thenthegraphofax2 is obtained by starting with the graph of x2, then flipping it over the x-axis, and then stretching or shrinking vertically by the positive number a. When a>0wesaythatthegraphof− ax2 “opens up”. When a<0wesay that the graph of ax2 “opens down”. I Cit i-a x-ax~S ~12 *************‘s-aXiS —10.? 148 2 If a, c, d and a = 0, then the graph of a(x + c) 2 + d is obtained by If a, c, d R and a = 0, then the graph of a(x + c)2 + d is obtained by 2 R 6 2 shiftingIf a, c, the d ⇥ graphR and ofaax=⇤ 2 0,horizontally then the graph by c, and of a vertically(x + c) + byd dis. obtained (Remember by shiftingshifting the the⇥ graph graph of of axax⇤ 2 horizontallyhorizontally by by cc,, and and vertically vertically by by dd.. (Remember (Remember thatthatd>d>0meansmovingup,0meansmovingup,d<d<0meansmovingdown,0meansmovingdown,c>c>0meansmoving0meansmoving thatleft,andd>c<0meansmovingup,0meansmovingd<right0meansmovingdown,.) c>0meansmoving leftleft,and,andc<c<0meansmoving0meansmovingrightright.).) 2 If a =0,thegraphofafunctionf(x)=a(x + c) 2+ d is called a parabola. If a =0,thegraphofafunctionf(x)=a(x + c)2 + d is called a parabola. 6 2 TheIf a point=0,thegraphofafunction⇤ ( c, d) 2 is called thefvertex(x)=aof(x the+ c parabola.) + d is called a parabola. The point⇤ ( c, d) R2 is called the vertex of the parabola.
    [Show full text]
  • Quadratic Equations
    Lecture 26 Quadratic Equations Quadraticpolynomials ................................ ....................... 2 Quadraticpolynomials ................................ ....................... 3 Quadraticequations and theirroots . .. .. .. .. .. .. .. .. .. ........................ 4 How to solve a binomial quadratic equation. ......................... 5 Solutioninsimplestradicalform ........................ ........................ 6 Quadraticequationswithnoroots........................ ....................... 7 Solving binomial equations by factoring. ........................... 8 Don’tloseroots!..................................... ...................... 9 Summary............................................. .................. 10 i Quadratic polynomials A quadratic polynomial is a polynomial of degree two. It can be written in the standard form ax2 + bx + c , where x is a variable, a, b, c are constants (numbers) and a = 0 . 6 The constants a, b, c are called the coefficients of the polynomial. Example 1 (quadratic polynomials). 4 4 3x2 + x ( a = 3, b = 1, c = ) − − 5 − −5 x2 ( a = 1, b = c = 0 ) x2 1 5x + √2 ( a = , b = 5, c = √2 ) 7 − 7 − 4x(x + 1) x (this is a quadratic polynomial which is not written in the standard form. − 2 Its standard form is 4x + 3x , where a = 4, b = 3, c = 0 ) 2 / 10 Quadratic polynomials Example 2 (polynomials, but not quadratic) x3 2x + 1 (this is a polynomial of degree 3 , not 2 ) − 3x 2 (this is a polynomial of degree 1 , not 2 ) − Example 3 (not polynomials) 2 1 1 x + x 2 + 1 , x are not polynomials − x A quadratic polynomial ax2 + bx + c is called sometimes a quadratic trinomial. A trinomial consists of three terms. Quadratic polynomials of type ax2 + bx or ax2 + c are called quadratic binomials. A binomial consists of two terms. Quadratic polynomials of type ax2 are called quadratic monomials. A monomial consists of one term. Quadratic polynomials (together with polynomials of degree 1 and 0 ) are the simplest polynomials.
    [Show full text]
  • Vocabulary Eighth Grade Math Module Four Linear Equations
    VOCABULARY EIGHTH GRADE MATH MODULE FOUR LINEAR EQUATIONS Coefficient: A number used to multiply a variable. Example: 6z means 6 times z, and "z" is a variable, so 6 is a coefficient. Equation: An equation is a statement about equality between two expressions. If the expression on the left side of the equal sign has the same value as the expression on the right side of the equal sign, then you have a true equation. Like Terms: Terms whose variables and their exponents are the same. Example: 7x and 2x are like terms because the variables are both “x”. But 7x and 7ͬͦ are NOT like terms because the x’s don’t have the same exponent. Linear Expression: An expression that contains a variable raised to the first power Solution: A solution of a linear equation in ͬ is a number, such that when all instances of ͬ are replaced with the number, the left side will equal the right side. Term: In Algebra a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs Unit Rate: A unit rate describes how many units of the first type of quantity correspond to one unit of the second type of quantity. Some common unit rates are miles per hour, cost per item & earnings per week Variable: A symbol for a number we don't know yet. It is usually a letter like x or y. Example: in x + 2 = 6, x is the variable. Average Speed: Average speed is found by taking the total distance traveled in a given time interval, divided by the time interval.
    [Show full text]
  • The Evolution of Equation-Solving: Linear, Quadratic, and Cubic
    California State University, San Bernardino CSUSB ScholarWorks Theses Digitization Project John M. Pfau Library 2006 The evolution of equation-solving: Linear, quadratic, and cubic Annabelle Louise Porter Follow this and additional works at: https://scholarworks.lib.csusb.edu/etd-project Part of the Mathematics Commons Recommended Citation Porter, Annabelle Louise, "The evolution of equation-solving: Linear, quadratic, and cubic" (2006). Theses Digitization Project. 3069. https://scholarworks.lib.csusb.edu/etd-project/3069 This Thesis is brought to you for free and open access by the John M. Pfau Library at CSUSB ScholarWorks. It has been accepted for inclusion in Theses Digitization Project by an authorized administrator of CSUSB ScholarWorks. For more information, please contact [email protected]. THE EVOLUTION OF EQUATION-SOLVING LINEAR, QUADRATIC, AND CUBIC A Project Presented to the Faculty of California State University, San Bernardino In Partial Fulfillment of the Requirements for the Degre Master of Arts in Teaching: Mathematics by Annabelle Louise Porter June 2006 THE EVOLUTION OF EQUATION-SOLVING: LINEAR, QUADRATIC, AND CUBIC A Project Presented to the Faculty of California State University, San Bernardino by Annabelle Louise Porter June 2006 Approved by: Shawnee McMurran, Committee Chair Date Laura Wallace, Committee Member , (Committee Member Peter Williams, Chair Davida Fischman Department of Mathematics MAT Coordinator Department of Mathematics ABSTRACT Algebra and algebraic thinking have been cornerstones of problem solving in many different cultures over time. Since ancient times, algebra has been used and developed in cultures around the world, and has undergone quite a bit of transformation. This paper is intended as a professional developmental tool to help secondary algebra teachers understand the concepts underlying the algorithms we use, how these algorithms developed, and why they work.
    [Show full text]
  • 2.3 Solving Linear Equations a Linear Equation Is an Equation in Which All Variables Are Raised to Only the 1 Power. in This
    2.3 Solving Linear Equations A linear equation is an equation in which all variables are raised to only the 1st power. In this section, we review the algebra of solving basic linear equations, see how these skills can be used in applied settings, look at linear equations with more than one variable, and then use this to find inverse functions of a linear function. Basic Linear Equations To solve a basic linear equation in one variable we use the basic operations of +, -, ´, ¸ to isolate the variable. Example 1 Solve each of the following: (a) 2x + 4 = 5x -12 (b) 3(2x - 4) = 2(x + 5) Applied Problems Example 2 Company A charges a monthly fee of $5.00 plus 99 cents per download for songs. Company B charges a monthly fee of $7.00 plus 79 cents per download for songs. (a) Give a formula for the monthly cost with each of these companies. (b) For what number of downloads will the monthly cost be the same for both companies? Example 3 A small business is considering hiring a new sales rep. to market its product in a nearby city. Two pay scales are under consideration. Pay Scale 1 Pay the sales rep. a base yearly salary of $10000 plus a commission of 8% of the total yearly sales. Pay Scale 2 Pay the sales rep a base yearly salary of $13000 plus a commission of 6% of the total yearly sales. (a) For each scale above, give a function formula to express the total yearly earnings as a function of the total yearly sales.
    [Show full text]