DOCSLIB.ORG

Nature of the Discriminant

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read 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. Which equation has imaginary roots? 7. How many x-intercepts does the graph of the 2 a. x2 – 9 = 0 function f(x) = x – 4x + 11 = 0 have? 2 b. x – x + 1 = 0 a. None 2 c. x + 2x + 1 = 0 b. One 2 d. x – 1 = 0 c. Two 2. What is the nature of the roots of 4x2 – 4x + 1 = 0? d. Cannot be determined a. Two real, rational solutions 8. Does the vertex of the graph of the function f(x) = b. Two real, irrational solutions x2 – 4x + 11 = 0 lie above, below, or on the x-axis? c. One real, rational solution a. Above d. Two imaginary/complex solutions b. Below 3. How many x-intercepts does the graph of the c. On function f(x) = x2 – 7x + 7 = 0 have? d. Cannot be determined a. None 9. How many solutions does 2x2 – 6x + 10 = 0 have? b. One c. Two d. Cannot be determined 4. Does the vertex of the graph of the function f(x) = 10. How many solutions does 3x2 – 6x – 10 = 0 have? x2 – 7x + 7 = 0 lie above, below, or on the x-axis? a. Above b. Below c. On 11. How many solutions does the function d. Cannot be determined x2 – 2x + 1 = 0 have? 5. How many x-intercepts does the graph of the function f(x) = x2 – 3x + 1 = 0 have? a. None b. One 12. Find all k such that 2x2 + kx + 3 = 0 has only one c. Two solution? Hint: the discriminant is k2 – 24; solve d. Cannot be determined k2 – 24 = 0. 6. Does the vertex of the graph of the function f(x) = x2 – 3x + 1 = 0 lie above, below, or on the x-axis? a. Above b. Below 13. Find all k such that x2 + kx + 2k = 0 has only one c. On solution? d. Cannot be determined .
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]
  • Elements of Chapter 9: Nonlinear Systems Examples
    Elements of Chapter 9: Nonlinear Systems To solve x0 = Ax, we use the ansatz that x(t) = eλtv. We found that λ is an eigenvalue of A, and v an associated eigenvector. We can also summarize the geometric behavior of the solutions by looking at a plot- However, there is an easier way to classify the stability of the origin (as an equilibrium), To find the eigenvalues, we compute the characteristic equation: p Tr(A) ± ∆ λ2 − Tr(A)λ + det(A) = 0 λ = 2 which depends on the discriminant ∆: • ∆ > 0: Real λ1; λ2. • ∆ < 0: Complex λ = a + ib • ∆ = 0: One eigenvalue. The type of solution depends on ∆, and in particular, where ∆ = 0: ∆ = 0 ) 0 = (Tr(A))2 − 4det(A) This is a parabola in the (Tr(A); det(A)) coordinate system, inside the parabola is where ∆ < 0 (complex roots), and outside the parabola is where ∆ > 0. We can then locate the position of our particular trace and determinant using the Poincar´eDiagram and it will tell us what the stability will be. Examples Given the system where x0 = Ax for each matrix A below, classify the origin using the Poincar´eDiagram: 1 −4 1. 4 −7 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = −6 Det(A) = −7 + 16 = 9 ∆ = 36 − 4 · 9 = 0 Therefore, we have a \degenerate sink" at the origin. 1 2 2. −5 −1 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = 0 Det(A) = −1 + 10 = 9 ∆ = 02 − 4 · 9 = −36 The origin is a center. 1 3. Given the system x0 = Ax where the matrix A depends on α, describe how the equilibrium solution changes depending on α (use the Poincar´e Diagram): 2 −5 (a) α −2 SOLUTION: The trace is 0, so that puts us on the \det(A)" axis.
    [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]
  • Polynomials and Quadratics
    Higher hsn .uk.net Mathematics UNIT 2 OUTCOME 1 Polynomials and Quadratics Contents Polynomials and Quadratics 64 1 Quadratics 64 2 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining the Equation of a Parabola 72 6 Solving Quadratic Inequalities 74 7 Intersections of Lines and Parabolas 76 8 Polynomials 77 9 Synthetic Division 78 10 Finding Unknown Coefficients 82 11 Finding Intersections of Curves 84 12 Determining the Equation of a Curve 86 13 Approximating Roots 88 HSN22100 This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/2.5/scotland/ Higher Mathematics Unit 2 – Polynomials and Quadratics OUTCOME 1 Polynomials and Quadratics 1 Quadratics A quadratic has the form ax2 + bx + c where a, b, and c are any real numbers, provided a ≠ 0 . You should already be familiar with the following. The graph of a quadratic is called a parabola . There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum This has a maximum turning point turning point To find the roots (i.e. solutions) of the quadratic equation ax2 + bx + c = 0, we can use: factorisation; completing the square (see Section 3); −b ± b2 − 4 ac the quadratic formula: x = (this is not given in the exam). 2a EXAMPLES 1. Find the roots of x2 −2 x − 3 = 0 .
    [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]
  • The Determinant and the Discriminant
    CHAPTER 2 The determinant and the discriminant In this chapter we discuss two indefinite quadratic forms: the determi- nant quadratic form det(a, b, c, d)=ad bc, − and the discriminant disc(a, b, c)=b2 4ac. − We will be interested in the integral representations of a given integer n by either of these, that is the set of solutions of the equations 4 ad bc = n, (a, b, c, d) Z − 2 and 2 3 b ac = n, (a, b, c) Z . − 2 For q either of these forms, we denote by Rq(n) the set of all such represen- tations. Consider the three basic questions of the previous chapter: (1) When is Rq(n) non-empty ? (2) If non-empty, how large Rq(n)is? (3) How is the set Rq(n) distributed as n varies ? In a suitable sense, a good portion of the answers to these question will be similar to the four and three square quadratic forms; but there will be major di↵erences coming from the fact that – det and disc are indefinite quadratic forms (have signature (2, 2) and (2, 1) over the reals), – det and disc admit isotropic vectors: there exist x Q4 (resp. Q3) such that det(x)=0(resp.disc(x) = 0). 2 1. Existence and number of representations by the determinant As the name suggest, determining Rdet(n) is equivalent to determining the integral 2 2 matrices of determinant n: ⇥ (n) ab Rdet(n) M (Z)= g = M2(Z), det(g)=n . ' 2 { cd 2 } ✓ ◆ n 0 Observe that the diagonal matrix a = has determinant n, and any 01 ✓ ◆ other matrix in the orbit SL2(Z).a is integral and has the same determinant.
    [Show full text]
  • Resultant and Discriminant of Polynomials
    RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results are well-known 19th century mathematics, but I have not inves- tigated the history, and no references are given. 1. Resultant n m Definition 1.1. Let f(x) = anx + ··· + a0 and g(x) = bmx + ··· + b0 be two polynomials of degrees (at most) n and m, respectively, with coefficients in an arbitrary field F . Their resultant R(f; g) = Rn;m(f; g) is the element of F given by the determinant of the (m + n) × (m + n) Sylvester matrix Syl(f; g) = Syln;m(f; g) given by 0an an−1 an−2 ::: 0 0 0 1 B 0 an an−1 ::: 0 0 0 C B . C B . C B . C B C B 0 0 0 : : : a1 a0 0 C B C B 0 0 0 : : : a2 a1 a0C B C (1.1) Bbm bm−1 bm−2 ::: 0 0 0 C B C B 0 bm bm−1 ::: 0 0 0 C B . C B . C B C @ 0 0 0 : : : b1 b0 0 A 0 0 0 : : : b2 b1 b0 where the m first rows contain the coefficients an; an−1; : : : ; a0 of f shifted 0; 1; : : : ; m − 1 steps and padded with zeros, and the n last rows contain the coefficients bm; bm−1; : : : ; b0 of g shifted 0; 1; : : : ; n−1 steps and padded with zeros. In other words, the entry at (i; j) equals an+i−j if 1 ≤ i ≤ m and bi−j if m + 1 ≤ i ≤ m + n, with ai = 0 if i > n or i < 0 and bi = 0 if i > m or i < 0.
    [Show full text]