DOCSLIB.ORG

Quadratic Functions, Parabolas, Problem Solving

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Quadratic Functions, Parabolas, Problem Solving pg097 [R] G1 5-36058 / HCG / Cannon & Elich cr 11-14-95 QC3 2.5 Quadratic Functions, Parabolas, and Problem Solving 97 58. If f~x! 5 2x 1 3 and the domain D of f is given by 63. Area Given the three lines y 5 m~x 1 4!, D 5 $x _ x 2 1 x 2 2 # 0%,then y 52m~x14!,andx52, for what value of m will (a) draw a graph of f,and thethreelinesformatrianglewithanareaof24? (b) find the range of f.(Hint: What values of y occur on 64. (a) Is the point ~21, 3! on the circle the graph?) x 2 1 y 2 2 4x 1 2y 2 20 5 0? 59. The length L of a metal rod is a linear function of its (b) If the answer is yes, find an equation for the line temperature T, where L is measured in centimeters and that is tangent to the circle at (21,3);iftheanswer T in degrees Celsius. The following measurements have is no, find the distance between the x-intercept been made: L 5 124.91 when T 5 0, and L 5 125.11 points of the circle. when T 5 100. 65. (a) Show that points A~1, Ï3! and B~2Ï3, 1! are on (a) Find a formula that gives L as a function of T. the circle x 2 1 y 2 5 4. (b) What is the length of the rod when its temperature (b) Find the area of the shaded region in the diagram. is 208 Celsius? (Hint: Show that segments OA and OB are perpen- (c) To what temperature should the rod be heated to dicular to each other. make it 125.17 cm long? y 60. Store Prices The owner of a grocery store finds that, A(1, 3) on average, the store can sell 872 gallons of milk per B(– 3, 1) week when the price per gallon is $1.98. When the price per gallon is $1.75, sales average 1125 gallons a week. Assume that the number N of gallons sold per week is x a linear function of the price P per gallon. O (a) Find a formula that gives N as a function of P. (b) If the price per gallon is $1.64, how many gallons should the store owner expect to sell per week? (c) To sell 1400 gallons per week, what price should the store set per gallon? 66. The horizontal line y 5 2 divides nABC into two re- 61. Equilateral Triangle Forwhatvalue(s)ofmwill the gions. Find the area of the two regions for the triangle triangle formed by the three lines y 522, y 5 mx14, with these vertices: and y 52mx 1 4 be equilateral? (a) A~0, 0!, B~4, 0!, C~0, 4! 62. Perimeter Find m such that the three lines y 5 mx16, (b) A~0, 0!, B~4, 0!, C~8, 4! y 52mx 1 6, and y 5 2formatrianglewitha 67. The vertices of nABC are A~0, 0!, B~4, 0!, C~0, 4!.If perimeter of 16. 0,k,4, then the horizontal line y 5 k will divide the triangle into two regions. Draw a diagram showing these regions for a typical k.Findthevalueofkfor which the areas of the two regions are equal. 2.5 QUADRATIC FUNCTIONS, PARABOLAS, AND PROBLEM SOLVING It is only fairly recently that the importance of nonlinearities has intruded itself into the world of the working scientist. Nonlinearity is one of those strange concepts that is defined by what it is not. As one physicist put it, “It is like having a zoo of nonelephants.” B. J. West Much of this course, and calculus courses to follow, deals with other inhabitants of the “zoo” of nonlinear functions, including families of polynomial, exponential, logarithmic, and trigonometric functions. All are important, but quadratic func- tions are among the simplest nonlinear functions mathematicians use to model the world. pg098 [V] G2 5-36058 / HCG / Cannon & Elich cr 10-31-95 QC 98 Chapter 2 Functions I had mathematical Definition: quadratic function curiosity very early. My A quadratic function is a function with an equation equivalent to father had in his library a 2 wonderful series of f ~x! 5 ax 1 bx 1 c, (1) German paperback books....OnewasEuler’s where a, b,andcare real numbers and a is not zero. Algebra. I discovered by myself how to solve For example, g~x! 5 5 2 4x 2 and h~x! 5 ~2x 2 1!2 2 x2 are quadratic func- equations. I remember that tions, but F~x! 5 Ïx 2 1 x 1 1andG~x!5~x23!2 2 x2 are not. (In fact, G is I did this by an incredible linear.) concentration and almost painful and not-quite Basic Transformations and Graphs of Quadratic Functions conscious effort. What I did amounted to completing The graphing techniques we introduced in Section 2.3 are collectively called basic the square in my head without paper or pencil. transformations. The graph of any linear function is a line, and we will show that Stan Ulam that graph of any quadratic function can be obtained from the core parabola, f ~x! 5 x 2, by applying basic transformations. We apply terminology from the core parabola to parabolas in general. The point (0, 0) is called the vertex of the core parabola, and the y-axis is the axis of symmetry. The axis of symmetry is a help in making a hand-sketch of a parabola. Whenever we locate a point of the parabola on one side of the axis of symmetry, we automatically have another point located symmetrically on the other side. We will derive a transformation form for a general quadratic function, an equation that identifies the vertex and axis of symmetry of the graph, but to graph any particular quadratic, you may not need all of the steps. Indeed, because a graphing calculator graphs any quadratic function, we could ask why we need the transformation form at all. Because a calculator graph is dependent on the window we choose and the pixel coordinates, we need an algebraic form from which we can read more exact information. To begin, we factor out the coefficient a from the x-terms,andthenaddand subtract the square of half the resulting x-coefficient to complete the square on x. b f~x! 5 ax2 1 bx 1 c 5 a x2 1 x 1 c S a D b b 2 b 2 5 a x 2 1 x 1 2 1 c FS a 4a 2D 4a 2G b 2 b 2 5 a x 1 1 c 2 . S 2aD S 4aD The final equation has the form f~x! 5 a~x 2 h!2 1 k (2) which we recognize as a core parabola shifted so that the vertex is at the point ~h, k! and the axis of symmetry is the line x 5 h. Parabola Features Looking at the derivation of Equation (2), we can make some observations about the graphs of quadratic functions. pg99 [R] G1 5-36058 / HCG / Cannon & Elich cr 11-14-95 QC4 2.5 Quadratic Functions, Parabolas, and Problem Solving 99 Graphs of quadratic functions For the quadratic function f~x! 5 ax2 1 bx 1 c: The graph is a parabola 2b Axis of with axis of symmetry x 5 . y a symmetry 2 The parabola opens upward if a . 0, downward if a , 0. 2b x =– b To find the coordinates of the vertex, set x 5 .Thenthey-coordinate is 2a 2a x 2b given by y 5 f . S 2aD x-intercept points The graph of every quadratic function intersects the y-axis (where x 5 0), but it need not have any x-intercept points. To find any x-intercepts, we solve the equation y-intercept Vertex f~x! 5 0. By its nature, every quadratic function has a maximum or a minimum point (depending on whether the parabola opens down or up) that occurs at the vertex of FIGURE 26 the parabola. See Figure 26. cEXAMPLE 1 Locating the vertex of a parabola The graph of the quad- ratic function is a parabola. Locate the vertex in two ways: (i) by writing the y b function in the form of Equation (2) and (ii) by setting x 52 . Sketch the 4 2a 3 graph. 2 2 2 (a) f ~x! 5 x 2 2x 2 3 (b) f ~x! 5 2x 1 4x 2 1 1 y = x2 Solution x 1 234 –4–3 –2 –1–1 (a) (i) Complete the square on the x-terms. –2 f ~x! 5 x 2 2 2x 2 3 5 ~x 2 2 2x 1 1! 2 1 2 3 –3 5 ~x 2 1!2 2 4, –4 (1, – 4) (a) f(x)=x2–2x–3 From this form, the graph is the core parabola shifted 1 unit right and 4 2 =(x–1) –4 units down; the vertex is at (1, 24). 2 b 22 y (ii) From f ~x! 5 x 2 2x 2 3, 2 2a 522·1 51. Substituting 1 for x, f ~1! 5 1 2 2·123524, so again the vertex is the point (1, 24). The 4 y = x2 graph is shown in Figure 27(a). 3 (b) (i) Before completing the square, we factor out 2 from the x-terms: 2 2 2 1 f ~x! 5 2x 1 4x 2 1 5 2~x 1 2x! 2 1 1 234 x 5 2~x 2 1 2x 1 1 2 1! 2 1 5 2~x 1 1!2 2 3.
Load more

Recommended publications
  • A Computational Approach to Solve a System of Transcendental Equations with Multi-Functions and Multi-Variables
    mathematics Article A Computational Approach to Solve a System of Transcendental Equations with Multi-Functions and Multi-Variables Chukwuma Ogbonnaya 1,2,* , Chamil Abeykoon 3 , Adel Nasser 1 and Ali Turan 4 1 Department of Mechanical, Aerospace and Civil Engineering, The University of Manchester, Manchester M13 9PL, UK; [email protected] 2 Faculty of Engineering and Technology, Alex Ekwueme Federal University, Ndufu Alike Ikwo, Abakaliki PMB 1010, Nigeria 3 Aerospace Research Institute and Northwest Composites Centre, School of Materials, The University of Manchester, Manchester M13 9PL, UK; [email protected] 4 Independent Researcher, Manchester M22 4ES, Lancashire, UK; [email protected] * Correspondence: [email protected]; Tel.: +44-(0)74-3850-3799 Abstract: A system of transcendental equations (SoTE) is a set of simultaneous equations containing at least a transcendental function. Solutions involving transcendental equations are often problematic, particularly in the form of a system of equations. This challenge has limited the number of equations, with inter-related multi-functions and multi-variables, often included in the mathematical modelling of physical systems during problem formulation. Here, we presented detailed steps for using a code- based modelling approach for solving SoTEs that may be encountered in science and engineering problems. A SoTE comprising six functions, including Sine-Gordon wave functions, was used to illustrate the steps. Parametric studies were performed to visualize how a change in the variables Citation: Ogbonnaya, C.; Abeykoon, affected the superposition of the waves as the independent variable varies from x1 = 1:0.0005:100 to C.; Nasser, A.; Turan, A.
    [Show full text]
  • Math 1232-04F (Survey of Calculus) Dr. J.S. Zheng Chapter R. Functions
    Math 1232-04F (Survey of Calculus) Dr. J.S. Zheng Chapter R. Functions, Graphs, and Models R.4 Slope and Linear Functions R.5* Nonlinear Functions and Models R.6 Exponential and Logarithmic Functions R.7* Mathematical Modeling and Curve Fitting • Linear Functions (11) Graph the following equations. Determine if they are functions. (a) y = 2 (b) x = 2 (c) y = 3x (d) y = −2x + 4 (12) Definition. The variable y is directly proportional to x (or varies directly with x) if there is some positive constant m such that y = mx. We call m the constant of proportionality, or variation constant. (13) The weight M of a person's muscles is directly proportional to the person's body weight W . It is known that a person weighing 200 lb has 80 lb of muscle. (a) Find an equation of variation expressing M as a function of W . (b) What is the muscle weight of a person weighing 120 lb? (14) Definition. A linear function is any function that can be written in the form y = mx + b or f(x) = mx + b, called the slope-intercept equation of a line. The constant m is called the slope. The point (0; b) is called the y-intercept. (15) Find the slope and y-intercept of the graph of 3x + 5y − 2 = 0. (16) Find an equation of the line that has slope 4 and passes through the point (−1; 1). (17) Definition. The equation y − y1 = m(x − x1) is called the point-slope equation of a line. The point is (x1; y1), and the slope is m.
    [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]
  • Factoring Polynomials
    EAP/GWL Rev. 1/2011 Page 1 of 5 Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. Example: — Set the factors of a polynomial equation (as opposed to an expression) equal to zero in order to solve for a variable: Example: To solve ,; , The flowchart below illustrates a sequence of steps for factoring polynomials. First, always factor out the Greatest Common Factor (GCF), if one exists. Is the equation a Binomial or a Trinomial? 1 Prime polynomials cannot be factored Yes No using integers alone. The Sum of Squares and the Four or more quadratic factors Special Cases? terms of the Sum and Difference of Binomial Trinomial Squares are (two terms) (three terms) Factor by Grouping: always Prime. 1. Group the terms with common factors and factor 1. Difference of Squares: out the GCF from each Perfe ct Square grouping. 1 , 3 Trinomial: 2. Sum of Squares: 1. 2. Continue factoring—by looking for Special Cases, 1 , 2 2. 3. Difference of Cubes: Grouping, etc.—until the 3 equation is in simplest form FYI: A Sum of Squares can 1 , 2 (or all factors are Prime). 4. Sum of Cubes: be factored using imaginary numbers if you rewrite it as a Difference of Squares: — 2 Use S.O.A.P to No Special √1 √1 Cases remember the signs for the factors of the 4 Completing the Square and the Quadratic Formula Sum and Difference Choose: of Cubes: are primarily methods for solving equations rather 1. Factor by Grouping than simply factoring expressions.
    [Show full text]
  • Write the Function in Standard Form
    Write The Function In Standard Form Bealle often suppurates featly when active Davidson lopper fleetly and ray her paedogenesis. Tressed Jesse still outmaneuvers: clinometric and georgic Augie diphthongises quite dirtily but mistitling her indumentum sustainedly. If undefended or gobioid Allen usually pulsate his Orientalism miming jauntily or blow-up stolidly and headfirst, how Alhambresque is Gustavo? Now the vertex always sits exactly smack dab between the roots, when you do have roots. For the two sides to be equal, the corresponding coefficients must be equal. So, changing the value of p vertically stretches or shrinks the parabola. To save problems you must sign in. This short tutorial helps you learn how to find vertex, focus, and directrix of a parabola equation with an example using the formulas. The draft was successfully published. To determine the domain and range of any function on a graph, the general idea is to assume that they are both real numbers, then look for places where no values exist. For our purposes, this is close enough. English has also become the most widely used second language. Simplify the radical, but notice that the number under the radical symbol is negative! On this lesson, you fill learn how to graph a quadratic function, find the axis of symmetry, vertex, and the x intercepts and y intercepts of a parabolawi. Be sure to write the terms with the exponent on the variable in descending order. Wendler Polynomial Webquest Introduction: By the end of this webquest, you will have a deeper understanding of polynomials. Anyone can ask a math question, and most questions get answers! Follow along with the highlighted text while you listen! And if I have an upward opening parabola, the vertex is going to be the minimum point.
    [Show full text]
  • Lesson 1: Multiplying and Factoring Polynomial Expressions
    NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M4 ALGEBRA I Lesson 1: Multiplying and Factoring Polynomial Expressions Classwork Opening Exercise Write expressions for the areas of the two rectangles in the figures given below. 8 2 2 Now write an expression for the area of this rectangle: 8 2 Example 1 The total area of this rectangle is represented by 3a + 3a. Find expressions for the dimensions of the total rectangle. 2 3 + 3 square units 2 푎 푎 Lesson 1: Multiplying and Factoring Polynomial Expressions Date: 2/2/14 S.1 This work is licensed under a © 2014 Common Core, Inc. Some rights reserved. commoncore.org Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M4 ALGEBRA I Exercises 1–3 Factor each by factoring out the Greatest Common Factor: 1. 10 + 5 푎푏 푎 2. 3 9 + 3 2 푔 ℎ − 푔 ℎ 12ℎ 3. 6 + 9 + 18 2 3 4 5 푦 푦 푦 Discussion: Language of Polynomials A prime number is a positive integer greater than 1 whose only positive integer factors are 1 and itself. A composite number is a positive integer greater than 1 that is not a prime number. A composite number can be written as the product of positive integers with at least one factor that is not 1 or itself. For example, the prime number 7 has only 1 and 7 as its factors. The composite number 6 has factors of 1, 2, 3, and 6; it could be written as the product 2 3.
    [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]
  • The Quadratic Formula You May Recall the Quadratic Formula for Roots of Quadratic Polynomials Ax2 + Bx + C
    For example, when we take the polynomial f (x) = x2 − 3x − 4, we obtain p 3 ± 9 + 16 2 which gives 4 and −1. Some quick terminology 2 I We say that 4 and −1 are roots of the polynomial x − 3x − 4 or solutions to the polynomial equation x2 − 3x − 4 = 0. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0. The quadratic formula You may recall the quadratic formula for roots of quadratic polynomials ax2 + bx + c. It says that the solutions to this polynomial are p −b ± b2 − 4ac : 2a Some quick terminology 2 I We say that 4 and −1 are roots of the polynomial x − 3x − 4 or solutions to the polynomial equation x2 − 3x − 4 = 0. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0. The quadratic formula You may recall the quadratic formula for roots of quadratic polynomials ax2 + bx + c. It says that the solutions to this polynomial are p −b ± b2 − 4ac : 2a For example, when we take the polynomial f (x) = x2 − 3x − 4, we obtain p 3 ± 9 + 16 2 which gives 4 and −1. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0.
    [Show full text]
  • Nature of the Discriminant
    Name: ___________________________ Date: ___________ Class Period: _____ Nature of the Discriminant Quadratic − b b 2 − 4ac x = b2 − 4ac Discriminant Formula 2a The discriminant predicts the “nature of the roots of a quadratic equation given that a, b, and c are rational numbers. It tells you the number of real roots/x-intercepts associated with a quadratic function. Value of the Example showing nature of roots of Graph indicating x-intercepts Discriminant b2 – 4ac ax2 + bx + c = 0 for y = ax2 + bx + c POSITIVE Not a perfect x2 – 2x – 7 = 0 2 b – 4ac > 0 square − (−2) (−2)2 − 4(1)(−7) x = 2(1) 2 32 2 4 2 x = = = 1 2 2 2 2 Discriminant: 32 There are two real roots. These roots are irrational. There are two x-intercepts. Perfect square x2 + 6x + 5 = 0 − 6 62 − 4(1)(5) x = 2(1) − 6 16 − 6 4 x = = = −1,−5 2 2 Discriminant: 16 There are two real roots. These roots are rational. There are two x-intercepts. ZERO b2 – 4ac = 0 x2 – 2x + 1 = 0 − (−2) (−2)2 − 4(1)(1) x = 2(1) 2 0 2 x = = = 1 2 2 Discriminant: 0 There is one real root (with a multiplicity of 2). This root is rational. There is one x-intercept. NEGATIVE b2 – 4ac < 0 x2 – 3x + 10 = 0 − (−3) (−3)2 − 4(1)(10) x = 2(1) 3 − 31 3 31 x = = i 2 2 2 Discriminant: -31 There are two complex/imaginary roots. There are no x-intercepts. Quadratic Formula and Discriminant Practice 1.
    [Show full text]
  • Formula Sheet 1 Factoring Formulas 2 Exponentiation Rules
    Formula Sheet 1 Factoring Formulas For any real numbers a and b, (a + b)2 = a2 + 2ab + b2 Square of a Sum (a − b)2 = a2 − 2ab + b2 Square of a Difference a2 − b2 = (a − b)(a + b) Difference of Squares a3 − b3 = (a − b)(a2 + ab + b2) Difference of Cubes a3 + b3 = (a + b)(a2 − ab + b2) Sum of Cubes 2 Exponentiation Rules p r For any real numbers a and b, and any rational numbers and , q s ap=qar=s = ap=q+r=s Product Rule ps+qr = a qs ap=q = ap=q−r=s Quotient Rule ar=s ps−qr = a qs (ap=q)r=s = apr=qs Power of a Power Rule (ab)p=q = ap=qbp=q Power of a Product Rule ap=q ap=q = Power of a Quotient Rule b bp=q a0 = 1 Zero Exponent 1 a−p=q = Negative Exponents ap=q 1 = ap=q Negative Exponents a−p=q Remember, there are different notations: p q a = a1=q p q ap = ap=q = (a1=q)p 1 3 Quadratic Formula Finally, the quadratic formula: if a, b and c are real numbers, then the quadratic polynomial equation ax2 + bx + c = 0 (3.1) has (either one or two) solutions p −b ± b2 − 4ac x = (3.2) 2a 4 Points and Lines Given two points in the plane, P = (x1; y1);Q = (x2; y2) you can obtain the following information: p 2 2 1. The distance between them, d(P; Q) = (x2 − x1) + (y2 − y1) . x + x y + y 2.
    [Show full text]
  • (Trying To) Solve Higher Order Polynomial Equations. Featuring a Recall of Polynomial Long Division
    (Trying to) solve Higher order polynomial equations. Featuring a recall of polynomial long division. Some history: The quadratic formula (Dating back to antiquity) allows us to solve any quadratic equation. ax2 + bx + c = 0 What about any cubic equation? ax3 + bx2 + cx + d = 0? In the 1540's Cardano produced a \cubic formula." It is too complicated to actually write down here. See today's extra credit. What about any quartic equation? ax4 + bx3 + cx2 + dx + e = 0? A few decades after Cardano, Ferrari produced a \quartic formula" More complicated still than the cubic formula. In the 1820's Galois (age ∼19) proved that there is no general algebraic formula for the solutions to a degree 5 polynomial. In fact there is no purely algebraic way to solve x5 − x − 1 = 0: Galois died in a duel at age 21. This means that, as disheartening as it may feel, we will never get a formulaic solution to a general polynomial equation. The best we can get is tricks that work sometimes. Think of these tricks as analogous to the strategies you use to factor a degree 2 polynomial. Trick 1 If you find one solution, then you can find a factor and reduce to a simpler polynomial equation. Example. 2x2 − x2 − 1 = 0 has x = 1 as a solution. This means that x − 1 MUST divide 2x3 − x2 − 1. Use polynomial long division to write 2x3 − x2 − 1 as (x − 1) · (something). Now find the remaining two roots 1 2 For you: Find all of the solutions to x3 + x2 + x + 1 = 0 given that x = −1 is a solution to this equation.
    [Show full text]
  • Calculus Method of Solving Quadratic Equations
    International Journal of Mathematics Research. ISSN 0976-5840 Volume 9, Number 2 (2017), pp. 155-159 © International Research Publication House http://www.irphouse.com Calculus Method of Solving Quadratic Equations Debjyoti Biswadev Sengupta#1 Smt. Sulochanadevi Singhania School, Jekegram, Thane(W), Maharashtra-400606, India. Abstract This paper introduces a new application of differential calculus and integral calculus to solve various forms of quadratic equations. Keywords: quadratic equations, differential calculus, integral calculus, infinitesimal integral calculus, constant of infinity. I. INTRODUCTION This paper is based on the usage of the principles of differential calculus and integral calculus to solve quadratic equations. It is hypothesized that if we use calculus to solve quadratic equations, we will arrive at the same solution that we may get from the quadratic formula method. It is also hypothesized that the quadratic formula can be derived from the quadratic formula. The paper involves the use of differential and integral calculus which takes solving of quadratic equations to the next level. The formulas which we will come across the paper, or use them, are shown in Table 1 under Heading IV. The paper shall be dealt under 2 headings. Heading II will deal with the use of differential calculus in solving quadratic equations while Heading III will deal with the use of integral calculus in quadratic equations, as far as solving of quadratic equations is concerned. Heading V is hugely concerned about the advantages of using calculus to solve quadratic equation. 156 Debjyoti Biswadev Sengupta II. USE OF DIFFERENTIAL CALCULUS TO SOLVE QUADRATIC EQUATIONS To solve quadratic equations, we shall require the use of first order differential calculus only.
    [Show full text]