SAJC Chapter 1 Equations Teachers 2021

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 1 of 54 (Pure Mathematics) Chapter 1: Equations Objectives At the end of the chapter, you should be able to: (a) solve quadratic equations using factorisation, completing the square, formula, sketching of graph and a GC; (b) understand and use the conditions for a quadra tic equation to have (i) two real and distinct roots, (ii) two real and equal roots, (iii) no real roots; (c) understand and use the conditions for a quadratic equation to be always positive (or always negative); (d) formulate a quadratic equation from a pro blem situation and interpret the solutio n in the context of the problem; (e) solve a pair of simultaneous equations, one linear and one quadratic, by substitution; (f) formulate a system of linear equations from a problem situation; (g) find the solution of a system of linear equations using a Graphing Calculator. Content 1.1 Quadratic Functions 1.1.1 Solving Quadratic Equations 1.1.2 Nature of Roots of a Quadratic Equation 1.1.3 Conditions for Quadratic Expression to be always positive (or always negative) 1.1.4 Intersection Problems Leading to Quadratic Equations 1.2 Simultaneous Linear and Quadratic Equations in Two Unknowns 1.2.1 Solving Simultaneous Linear and Quadratic Equations in Two Unknowns 1.2.2 Application questions 1.3 Systems of linear equation (SOLE) 1.3.1 Using Graphing Calculator to solve SOLE 1.3.2 Modelling a System of Linear Equations to Solve Practical Problems References 1. New Additional Mathematics, Ho Soo Thong (Msc, Dip Ed), Khor Nyak Hiong (Bsc, Dip Ed) 2. New Syllabus Additional Mathematics (7th Edition), Shinglee Publishers Ptd Ltd

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 2 of 54 1.1 Quadratic Functions The expression   2f x ax bx c   , where 0a , is called a quadratic function. When the graph of the function 2y ax bx c   is drawn, two types of graphs are obtained, depending on the value of a. 0a U shape 0a Inverted – U shape Note: 1. a is the coefficient of 2x b is the coefficient of x c is the coefficient of 0x (also known as the constant term or the term independent of x) 2. If 0a , the curve has a minimum point at X. For example, in 22 7 4y x x   , 2a , 7b , 4c . Since 20a , the curve is U shape and has a minimum point. If 0a , the curve has a maximum point at Y. For example, in 2 69y x x   , 1a , 6b , 9c . Since 10a  , the curve is Inverted – U shape and has a maximum point. 3. The shape of a quadratic function is symmetrical about its minimum or maximum point. 1.1.1 Solving Quadratic Equations The general form of a quadratic equation is 2 0ax bx c   , 0a . There are five methods of solving a quadratic equation, namely: (a) Factorisation (b) Completing the Square (c) Formula (d) Graphical Method (e) GC X Y

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 3 of 54 (a) Factorisation Method 1. Bring all the terms to one side of the equation such that the other side is 0. 2. Use Cross Factorisation Method to factorize the equation. Example 1 Solve the quadratic equation 23 2 0xx   using factorization. Solution: 23 2 0xx   (3 2)( 1) 0 3 2 0 or 1 0 2 or 13 xx xx xx          Note: 1. The solution of the quadratic equation, i.e. 2 or 13xx  are also called the roots of the quadratic equation. 2. (3 2) and ( 1)xx are called factors of the quadratic expression 232xx . (b) Completing the Square A quadratic equation of the form 2 0ax bx c   , 0a can be expressed to the form   2 0a x p q   by completing the square. Note: The coefficient of 2x must be 1 before completing the square method can be carried out. Example 2 Solve the quadratic equation 22 8 1 0xx   by completing the square. Solution:         2 2 2 2 2 2 2 2 2 8 1 0 2 4 1 0 2 4 2 2 1 0 2 2 2 1 0 2 2 7 xx xx xx x x                  72 2 72 2 x x     True or False?   3 2 1 2 3 2 2 or 1 2 xx xx         Ans: False

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 4 of 54 (c) Formula The general formula of solving a quadratic equation of the form 2 0ax bx c   , 0a is 2 4 2 b b acx a    Example 3 Solve the quadratic equation 2 7 3 0xx   using formula. Solution: 2 7 3 0xx   27 7 4(1)( 3) 2 7 61 2 7 61 7 61 or 22 x          Note: The expression 2 4b ac in the general formula is known as the discriminant of the quadratic equation as it determines the nature (type) of roots that a quadratic equation has. (d) Graphical Method Example 4 Solve the quadratic equation 2 5 6 0xx   using graphical method. Solution: From the graph, 2 or 3xx . Note: The x – intercepts of the curve 2 56y x x   are the solutions of the quadratic equation 2 5 6 0xx   . 3 y x 2 6 2 56y x x   Why are the x-intercepts the solution for 2 5 6 0xx   ? Ans: The x-intercepts are the points of intersection between the line 0y and the curve 2 56y x x   .

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 5 of 54 (e) GC Example 5 Solve the quadratic equation 27 6 5 0xx   using a calculator. Solution: Steps Screenshot Remarks Press Select PlySmlt2 Press À to select POLYNOMIAL ROOT FINDER Select 2 in the ORDER option Press s (NEXT) to go to the next screen ORDER prefers to the order of the polynomial that you are solving. (i.e. highest degree of the polynomial) Enter the coefficients of the quadratic equation (i.e. a, b and c) and select “  ” or “  ” according to the equation that you are solving. Press s (SOLVE) to solve the equation Note: Press F D will sometimes convert decimal to fraction The solutions are given by 1x and 2x . From GC, 0.519 or 1.38xx  Note: GC may not give exact values. If the question requires exact answer, do not use GC but you can always use GC to check your answer.

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 6 of 54 Exercise 1 1. Solve the following equations using the stated method in the bracket. (a) 22 11 12xx  (factorisation) (b) 23 8 1 0xx   (completing the square) (c) 2 6 1 0xx    (formula) (d) 22 13 14 0xx   (GC) Solution: (a)    2 2 2 11 12 2 11 12 0 2 3 4 0 2 3 0 or 4 0 3 or 42 xx xx xx xx xx              (b) 23 8 1 0xx   2 83 1 0 3xx    22 2 8 4 43 1 0 3 3 3xx                   2 2 4 163 1 0 33 4 13 39 4 13 39 4 13 3 x x x x            (c) 2 6 1 0xx    26 ( 6) 4( 1)(1) 2( 1) 6 40 2 6 2 10 2 3 10 or 3 10 x x x xx              (d) Using GC, 5.14 or 1.36xx Answer 1 (a) 11 or 42xx , (b) 4 13 3x  (c) (3 10)x  (d) 5.14 or 1.36xx

SAJC 2021 H1 Mathematics Chapter 1: Equations Page 7 of 54 1.1.2 Nature of Roots of a Quadratic Equation Recall that the solutions of a quadratic equation 2 0ax bx c   , 0a are given by 2 4 2 b b acx a    . The expression 2 4b ac in the general formula is known as the discriminant of the equation as it determines the nature (type) of roots that a quadratic equation has. Learning Experience Purpose: To understand the connection between the discriminant and the nature of roots of an equation and to develop reasoning skills. (1) Complete the table below. Quadratic equation Discri minant 2 4b ac Quadratic formula to determine the roots 2 4 2 b b acx a    Roots Nature of roots Sketc