Revision and final examination preparation – Week 1 focus
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Subject: Mathematics
Class: Grade 12
Term: Term 4
Week: 1
Theme: General lesson support
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This week marks the beginning of our focused revision for the final Grade 12 Mathematics examination. Mathematics is a fundamental subject, not just for academic progression, but for understanding and navigating the world around us. From budgeting and financial planning (essential for every South African household) to interpreting statistics related to national issues like employment and healthcare, mathematical literacy is crucial.
Furthermore, a strong foundation in mathematics opens doors to a vast range of career opportunities in fields like engineering, finance, technology, and even the arts.
2.1 Quadratic Equations A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants and a ≠
0. Methods of Solving: Factorization: This method involves expressing the quadratic expression as a product of two linear factors. It's efficient when the quadratic expression can be easily factored.
Completing the Square: This method transforms the quadratic equation into the form (x + p)² = q, allowing you to solve for x. It's particularly useful when factorization is not straightforward.
Quadratic Formula: This formula provides a general solution for any quadratic equation: x = (-b ± √(b² - 4ac)) / (2a). It's reliable but can be computationally intensive. The discriminant, b² - 4ac, determines the nature of the roots: b² - 4ac > 0: Two distinct real roots. b² - 4ac = 0: One real root (repeated root). b² - 4ac 0 or ax² + bx + c ≤
0. Solving Quadratic Inequalities: Rewrite the inequality in the standard form (e.g., ax² + bx + c > 0). Solve the corresponding quadratic equation ax² + bx + c = 0 to find the critical values (roots). Sketch a rough graph of the quadratic function y = ax² + bx + c. The critical values are the x-intercepts. Determine the intervals where the quadratic function satisfies the inequality. Express the solution in interval notation.
Worked Example 4: Solve x² - 3x - 10 d = 2 Substitute d = 2 into equation 1: a + 2(2) = 7 => a = 3 Therefore, the first term is 3 and the common difference is 2.