Chapter 2 – Algebra II 02 Learning Outcomes In this chapter you have learned to: Solve linear equations Solve simultaneous linear equations in two unknowns Solve simultaneous linear equations in three unknowns Solve quadratic equations of the form ax2 + bx + c = 0 Solve simultaneous equations: one linear, one non-linear Use the factor theorem for polynomials Solve cubic equations Manipulate formulae Determine unknown coefficients Solve problems using linear equations Solve problems using simultaneous equations Solve problems using quadratic equations Solve problems using cubic equations

02 Algebra II Solving Linear Equations Linear equations can be solved: (a) Using Trial and Error: 3(x – 1) = 9 3(1 – 1) ≠ 9 3(2 – 1) ≠ 9 3(3 – 1) ≠ 9 3(4 – 1) = 9 Solution x = 4 Solve for x if 3(x – 1) = 9 (b) Using Graphs: Solve for x if 2x + 4 = 3x + 3 Using the same axes and scales, graph y = 2x + 4 and y = 3x + 3. Solution x = 1 (c) Using Algebra: Solve for a if 3(4a – 6) + 25(a + 2) = a – 4 12a – 18 + 25a + 50 = a – 4 37a + 32 = a – 4 36a = −36 Solution a = −1

Solving Simultaneous Linear Equations in Two Variables 02 Algebra II Solving Simultaneous Linear Equations in Two Variables Simultaneous linear equations in two unknowns can be solved: (a) Using Substitution: 7x – 8y = –1 and x – 4y = –23 Solve for x and y if From the 2nd equation x = 4y –23 Substitute this into the 1st equation gives 7(4y – 23) – 8y = –1 28y – 161 – 8y = –1 20y = 160 y = 8 x = 4(8) –23 x = 32–23 x = 9 (b) Using Graphs: Using the same axes and scales, gives Solution x = 9, y = 8 Solve for x and y if 7x – 8y = –1 and x – 4y = –23 (c) Using Elimination: Solve for x and y if 7x – 8y = –1 and x – 4y = –23 7x – 8y = –1 x – 4y = –23 (multiply by −2) −2x + 8y = 46 (adding) 5x = 45 x = 9 7(9) – 8y = –1 63 – 8y = –1 8y = 64 y = 8 Solution x = 9, y = 8

Solving Simultaneous Equations in Three Variables 02 Algebra II Solving Simultaneous Equations in Three Variables Solve the equations: x + y + z = 6 5x + 3y – 2z = 5 3x – 7y + z = –8 The variable z looks easiest to eliminate. Label the equations: x + y + z = 6 Equation A 5x + 3y – 2z = 5 Equation B 3x – 7y + z = –8 Equation C Multiply Equation A by 2. Add: 2x + 2y + 2z = 12 7x + 5y = 17 Equation 1 Now use these x + y + z = 6 Equation A 3x − 7y + z = −8 Equation C Multiply Equation C by −1. Add: x + y + z = 6 −3x + 7y –z = 8 −2x + 8y = 14 Equation 2 Now solve between Equation 1 and 2 7x + 5y = 17 Equation 1 66y = 132 14x + 10y = 34 (Multiply E1 by 2) −14x + 56y = 98 (Multiply E2 by 7) y = 2 Substitution into Equation 1 gives x = 1 Substitution of x and y into Equation A gives z = 3

Solving Quadratic Equations 02 Algebra II Solving Quadratic Equations Quadratic equations of the form ax2 + bx + c = 0 can be solved: (a) Using Factors (if factorisable): Solve 3x2 + 5x – 12 = 0 Factorise 3x2 + 5x – 12 (3x – 4)(x + 3) (3x – 4)(x + 3) =0 𝑥= 4 3 or 𝑥=−3 (b) Using Graphs: To estimate the roots (solutions), Draw the graph and read off the values of x where the graph intersects the x-axis. The graph of the function y = 2x2 + 5x – 9 is shown. Estimate the roots of this equation. Solution 𝑥≈−3.7 or 𝑥≈1.2 (c) Using the Quadratic Formula: Using the quadratic Formula to solve x2 – x – 6 = 0

Simultaneous Equations: One Linear, One Non-Linear 02 Algebra II Simultaneous Equations: One Linear, One Non-Linear Solve for x and y if x + 4y = 1 and 2x2 + 3xy = 35 2x2 + 3xy = 35 x + 4y = 1 From the linear equation x = 1 − 4y Substitute this value for x into the other equation. 2(1 − 4y)2 + 3(1 − 4y)y = 35 2(1 + 16y2 −8y) + 3y − 12y2 = 35 2 + 32y2 − 16y + 3y − 12y2 = 35 20y2 − 13y − 33 = 0 (20y – 33)(y + 1) = 0 𝑦= 33 20 or 𝑦=−1 Substitute these values into 𝑥=1−4𝑦 to find the values for 𝑥. ⟹𝑥=− 28 5 or 𝑥=5

02 Algebra II The Factor Theorem A polynomial f(x) has a factor (x – a) if and only if f(a) = 0. The factor theorem can be very useful when finding the roots or factors of cubic polynomials We can use the factor theorem to find one factor and then use long division to proceed further. Solve the equation 2x3 – 3x2 – 17x + 30 = 0. By Trial and Error x = −3 is a root therefore using the Factor Theorem x + 3 is a factor. We can now use long division to find the other factors and, hence, the other roots. ⇒ (x + 3)(2x2 – 9x + 10) = 0 (x + 3)(2x – 5)(x – 2) = 0 ⟹𝑥=−3, 𝑥= 5 2 𝑥=2 are the solutions

Using a Graph to Estimate a Polynomial 02 Algebra II Using a Graph to Estimate a Polynomial The values of x for which f(x) = 0 are called the roots or zeros of the function. The degree of a polynomial is the highest power within the polynomial. The maximum number of distinct roots a polynomial can have is the same as its degree. The leading coefficient of a polynomial is the coefficient of the term with the highest power. The end behaviour of a polynomial function will always follow these rules: If the polynomial is of even degree, then the arms of the graph both point up or both point down. If the polynomial is of odd degree, then one arm points up and one points down. If the leading coefficient is positive, then the right arm points up. If the leading coefficient is negative, then the right arm points down. Example: 𝑓(𝑥)=𝑥3−6𝑥2+11𝑥−6 Degree is 3, which is odd; leading coefficient is 1, which is positive. ∴ The left arm points down and the right arm points up.

02 Algebra II Find a polynomial of degree 5 whose graph is shown below. The roots are 𝑥=−4, 𝑥=0, 𝑥=2, 𝑥=5, 𝑎𝑛𝑑 𝑥=5 Two possible answers are: 𝑥+4 𝑥 𝑥−2 𝑥−5 𝑥−5 OR − 𝑥+4 𝑥 𝑥−2 𝑥−5 𝑥−5 As the right arm points down the leading coefficient must be negative so the function is 𝑓 𝑥 =− 𝑥+4 𝑥 𝑥−2 𝑥−5 𝑥−5 𝑓 𝑥 =− 𝑥+4 𝑥 𝑥−2 (𝑥−5) 2

Manipulation of Formulae 02 Algebra II Manipulation of Formulae The volume of a cylinder is given by the formula 𝑉=𝜋 𝑟 2 ℎ However, if we wish to find the radius of a cylinder using this formula, we must rearrange or manipulate the formula: 𝑉=𝜋 𝑟 2 ℎ 𝑟 2 = 𝑉 𝜋ℎ 𝑟= 𝑉 𝜋ℎ

02 Algebra II Unknown Coefficients If c(x – a)2 + b = 3x2 – 6x + 5 for all values of x, find the values of a, b, c ∈ R. If x2 + px + r is a factor of x3 + 2px2 + 9x + 2r, find the value of p and r. Let (x + k) be the other factor. c(x – a)2 + b = 3x2 – 6x + 5 (x + k)(x2 + px + r) = x3 + 2px2 + 9x + 2r c(x2 – 2ax + a2) + b = 3x2 – 6x + 5 x3 + px2 + rx + kx2 + kpx + kr = x3 + 2px2 + 9x + 2r cx2 – 2acx + a2c + b = 3x2 – 6x + 5 x3 + px2 + kx2 + rx + kpx + kr = x3 + 2px2 + 9x + 2r We now group the like terms: We can now equate like terms from each side: x3 terms x3 = x3 x2 terms px2 + kx2 = 2px2 p + k = 2p k = p x terms rx + kpx = 9x r + kp = 9 Constants kr = 2 r x2 terms cx2 = 3x2 c = 3 x terms – 2acx = – 6x – 2ac = – 6 ac = 3 a(3) = 3 a = 1 Constants a2c + b = 5 (1)2(3) + b = 5 3 + b = 5 b = 2 k = 2 As k = p p = 2 As r + kp = 9 r + (2)(2) = 9 r = 5

Problem Solving Using Algebra 02 Algebra II Problem Solving Using Algebra All types of mathematical problems can be solved using: linear equations simultaneous equations quadratic equations cubic equations