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HKDSE Mathematics Ronald Hui Tak Sun Secondary School
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Missing Homework Summer Holiday Homework 1 1, 10, 14 Summer Holiday Homework 2 1, 9, 14 SHW1-R1 9, 10 22 October 2015 Ronald HUI
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Missing Homework SHW2-A1 10 SHW2-B1 9 SHW2-C1 8, 9, 10, 12, 13, 14, 20 RE2 9 22 October 2015 Ronald HUI
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Missing Homework SHW2-R1 I got 8 only! SHW2-P1 I got 4 only!!! SHW3-01, 3-A1, 3-B1 Today!!!! 22 October 2015 Ronald HUI
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Summary on “AND” 22 October 2015 Ronald HUI
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Summary on “OR” 22 October 2015 Ronald HUI
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Book 5A Chapter 3 Solving Quadratic Inequalities in One Unknown by the Graphical Method
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If an inequality contains only one unknown and the highest degree of the terms is 2, the inequality is called a quadratic inequality in one unknown. Example: (i) x 2 + 3x – 10 > 0 (ii) x 2 – 2x + 1 0 (iv) –2x 2 + x + 1 0 (iii) x 2 + 3x – 10 < 0
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We have to find all values of x satisfying the quadratic inequality. How can we solve this kind of quadratic inequalities?
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Let’s see how to solve x 2 + 2x – 3 > 0 by the graphical method.
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Solve the quadratic inequality x 2 + 2x – 3 > 0 graphically. Consider the graph of the quadratic function y = x 2 + 2x – 3. Points above the x-axis Points below the x-axis y > 0 y < 0
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Solve the quadratic inequality x 2 + 2x – 3 > 0 graphically. Consider the graph of the quadratic function y = x 2 + 2x – 3. For x < –3: The graph is above the x-axis. For –3 < x < 1: The graph is below the x-axis. For x > 1: The graph is above the x-axis. y > 0 y < 0 y > 0
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For y > 0 (i.e. x 2 + 2x – 3 > 0), ∴ The solutions of x 2 + 2x – 3 > 0 are x 1. For x < –3: y > 0 For –3 < x < 1: y < 0 For x > 1: y > 0 Solve the quadratic inequality x 2 + 2x – 3 > 0 graphically. Consider the graph of the quadratic function y = x 2 + 2x – 3. For y > 0 (i.e. x 2 + 2x – 3 > 0), the range of values of x are x 1.
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Solve the quadratic inequality x 2 + 2x – 3 > 0 graphically. Consider the graph of the quadratic function y = x 2 + 2x – 3. Using the above graph, we can also find the solutions of x 2 + 2x – 3 0. For x 0 For –3 < x < 1: y < 0 For x > 1: y > 0
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Solve the quadratic inequality x 2 + 2x – 3 0 graphically. For x –3: The graph is on or above the x-axis. y 0 For –3 x 1: The graph is on or below the x-axis. y 0 For x 1: The graph is on or above the x-axis. y 0
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the range of values of x are –3 x 1. Solve the quadratic inequality x 2 + 2x – 3 0 graphically. Note: In fact, the inequalities x 2 + 2x – 3 0 and x 2 + 2x – 3 < 0 can also be solved using the graph above. ∴ The solutions of x 2 + 2x – 3 0 are –3 x 1. For y 0 (i.e. x 2 + 2x – 3 0), For x –3: y 0 For –3 x 1: y 0 For x 1: y 0
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Let’s see how we can solve the quadratic inequality x 2 + 3x – 10 < 0 without any given graph.
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Consider the corresponding quadratic function Solve the quadratic inequality x 2 + 3x – 10 < 0 graphically. Step 1 Write down the corresponding quadratic function y = ax 2 + bx + c. y = x 2 + 3x – 10.
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Step 2 Find the x-intercept(s) and the direction of opening of the graph of the quadratic function. Then, sketch the graph. Solve the quadratic inequality x 2 + 3x – 10 < 0 graphically. Sketch the graph of y = x 2 + 3x – 10. x 2 + 3x – 10 = 0 (x + 5)(x – 2) = 0 x = –5 or x = 2 When y = 0, ∴ The x-intercepts of the graph are –5 and 2.
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Step 2 Find the x-intercept(s) and the direction of opening of the graph of the quadratic function. Then, sketch the graph. Solve the quadratic inequality x 2 + 3x – 10 < 0 graphically. Sketch the graph of y = x 2 + 3x – 10. ∵ The coefficient of x 2 is 1 (> 0). ∴ The graph opens upwards. y = x 2 + 3x – 10
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Read the solutions of the quadratic inequality from the graph. Step 3 y = x 2 + 3x – 10 For y < 0 (i.e. x 2 + 3x – 10 < 0), the corresponding part of the graph is below the x-axis. Solve the quadratic inequality x 2 + 3x – 10 < 0 graphically.
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y = x 2 + 3x – 10 ∵ The graph of y = x 2 + 3x – 10 is below the x-axis when –5 < x < 2. ∴ The solutions of x 2 + 3x – 10 < 0 are –5 < x < 2. Solve the quadratic inequality x 2 + 3x – 10 < 0 graphically. Read the solutions of the quadratic inequality from the graph. Step 3
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Follow-up question Solve the quadratic inequality –2x 2 + x + 1 0 graphically. Consider the corresponding quadratic function y = –2x 2 + x + 1. When y = 0, –2x 2 + x + 1 = 0 2x 2 – x – 1 = 0 (2x + 1)(x – 1) = 0 2 1 x = – or x = 1 ∴ The x-intercepts of the graph of y = –2x 2 + x + 1 are – and 1. 2 1
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∵ The coefficient of x 2 is –2 (< 0). ∴ The graph of y = –2x 2 + x + 1 opens downwards. Sketch the graph of y = –2x 2 + x + 1: From the graph, the solutions of –2x 2 + x + 1 0 are x – or x 1. 2 1
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