Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Factorize algebraic expressions.
• Solve quadratic equations using factorization.
• Solve simultaneous linear equations.
• Apply the Theorem of Pythagoras in right-angled triangles.

Lesson summary

This week's focus is on revising essential concepts covered thus far in Grade 10 Mathematics. We are consolidating our understanding to build a strong foundation for future, more complex topics. A solid grasp of these fundamentals is crucial not just for academic success, but also for tackling everyday situations that require mathematical reasoning. For instance, understanding financial calculations (like interest rates) is directly applicable to managing personal finances or understanding loan agreements. Similarly, spatial reasoning, which is tied to geometry and transformations, is important for interpreting maps and blueprints, helpful in construction or engineering fields.

Lesson notes

2.1 Factorization of Algebraic Expressions Factorization is the process of expressing an algebraic expression as a product of its factors. This is the reverse of expansion.

Several techniques are used: Common Factor: Identify the highest common factor (HCF) of all terms and factor it out.

Example: Factorize `6x² + 9x`. The HCF of 6 and 9 is 3, and the HCF of x² and x is x.

Therefore: `6x² + 9x = 3x(2x + 3)` Difference of Two Squares: Applicable when the expression is in the form `a² - b²`. The factorization is `(a + b)(a - b)`.

Example: Factorize `x² - 16`. Here, a = x and b =

4. Therefore: `x² - 16 = (x + 4)(x - 4)` Trinomials: Simple Trinomials (x² + bx + c): Find two numbers that add up to 'b' and multiply to 'c'.

Example: Factorize `x² + 5x + 6`. We need two numbers that add to 5 and multiply to

6. These are 2 and

3. Therefore: `x² + 5x + 6 = (x + 2)(x + 3)` Complex Trinomials (ax² + bx + c): This is more involved. You can use the "ac method" or trial and error. In the ac method, you find two numbers that add up to 'b' and multiply to 'ac'.

Example: Factorize `2x² + 7x + 3`. a=2, b=7, c=

3. So, ac =

6. We need two numbers that add to 7 and multiply to

6. These are 1 and

6. Rewrite the middle term: `2x² + x + 6x + 3` Factor by grouping: `x(2x + 1) + 3(2x + 1)` Therefore: `2x² + 7x + 3 = (2x + 1)(x + 3)` 2.2 Solving Quadratic Equations A quadratic equation is an equation of the form `ax² + bx + c = 0`, where a ≠

0. Factorization Method: If the quadratic expression can be factorized, set each factor equal to zero and solve for x.

Example: Solve `x² - 4x + 3 = 0`. Factorizing, we get `(x - 1)(x - 3) = 0`.

Therefore, `x - 1 = 0` or `x - 3 = 0`. So, `x = 1` or `x = 3`.

Quadratic Formula: When factorization is difficult or impossible, use the quadratic formula: `x = (-b ± √(b² - 4ac)) / 2a`

Example: Solve `2x² + 5x - 3 = 0`. Here, a = 2, b = 5, and c = -3. `x = (-5 ± √(5² - 4 2 -3)) / (2 * 2)` `x = (-5 ± √(25 + 24)) / 4` `x = (-5 ± √49) / 4` `x = (-5 ± 7) / 4` Therefore, `x = (-5 + 7) / 4 = 1/2` or `x = (-5 - 7) / 4 = -3` 2.3 Simplifying Algebraic Fractions Simplifying algebraic fractions involves factoring the numerator and denominator and then cancelling any common factors.

Example: Simplify: `(x² - 4) / (x² + 4x + 4)` Factorize the numerator (difference of two squares): `(x + 2)(x - 2)` Factorize the denominator (perfect square trinomial): `(x + 2)(x + 2)` Therefore: `(x² - 4) / (x² + 4x + 4) = [(x + 2)(x - 2)] / [(x + 2)(x + 2)] = (x - 2) / (x + 2)` 2.4 Solving Simultaneous Equations These are two or more equations with the same variables. We aim to find values for the variables that satisfy all equations.

Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation.

Example: Solve: `x + y = 5` and `2x - y = 1` From the first equation, `x = 5 - y`.

Substitute this into the second equation: `2(5 - y) - y = 1`.

Simplify: `10 - 2y - y = 1` `-3y = -9` `y = 3` Substitute y = 3 back into `x = 5 - y`: `x = 5 - 3 = 2`.

Therefore, the solution is x = 2 and y =

3. Elimination Method: Multiply one or both equations by constants so that the coefficients of one of the variables are opposites. Then add the equations together to eliminate that variable.

Using the same example: `x + y = 5` and `2x - y = 1` Notice that the y coefficients are already opposites. Adding the equations directly eliminates y: `(x + y) + (2x - y) = 5 + 1` `3x = 6` `x = 2` Substitute x = 2 back into `x + y = 5`: `2 + y = 5`, so `y = 3`.

Therefore, the solution is x = 2 and y = 3. 2.5 Theorem of Pythagoras In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. `a² + b² = c²`, where c is the hypotenuse.

Example: A right-angled triangle has sides of length 3cm and 4cm. Find the length of the hypotenuse. `3² + 4² = c²` `9 + 16 = c²` `25 = c²` `c = √25 = 5 cm` 2.6 Angle Properties of Triangles Angle Sum of a Triangle: The sum of the interior angles of any triangle is always 180°.

Exterior Angle of a Triangle: An exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Isosceles Triangle: Two sides are equal and the angles opposite these sides are also equal.

Equilateral Triangle: All three sides are equal, and all three angles are 60°.

Angles formed by parallel lines: Corresponding angles are equal, alternate angles are equal and co-interior angles add up to 180 degrees. Guided Practice (With Solutions)

Question 1: Factorize completely: `4x² - 9` Solution: This is a difference of two squares. `4x² = (2x)²` and `9 = 3²`.

Therefore, `4x² - 9 = (2x + 3)(2x - 3)`

Commentary: Identifying the structure (difference of two squares) is crucial.

Question 2: Solve for x: `x² - 5x + 6 = 0` Solution: We need two numbers that add to -5 and multiply to

6. These are -2 and -3. `(x - 2)(x - 3) = 0` Therefore, `x - 2 = 0` or `x - 3 = 0`. So, `x = 2` or `x = 3`.