Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Solve quadratic equations using three different methods.
• Solve quadratic inequalities and show solutions on a number line.
• Understand the 'nature of roots' using the discriminant.
• Identify key features for sketching a quadratic function.

Lesson summary

This week, we delve deeper into the world of equations and inequalities. Building on your prior knowledge of solving linear equations, we'll focus on quadratic equations and inequalities, which are prevalent in various real-world scenarios. Understanding these concepts is crucial, not only for your mathematical development but also for problem-solving in fields like engineering, finance, and even everyday situations. Imagine calculating the optimal dimensions for a vegetable garden in your backyard to maximize yield, or understanding loan interest calculations – these often involve quadratic equations and inequalities.

Lesson notes

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. The solutions to the quadratic equation are called roots or zeros. 2.1.1 Solving by Factorization: This method involves expressing the quadratic expression as a product of two linear factors.

Process: Rearrange the equation into the standard form ax² + bx + c =

0. Factorize the quadratic expression ax² + bx + c.

Think: "What two numbers multiply to give 'ac' and add up to 'b'?" Set each factor equal to zero. Solve each linear equation to find the values of x.

Example 1: Solve x² + 5x + 6 = 0 The equation is already in standard form.

Factorize: x² + 5x + 6 = (x + 2)(x + 3) (Because 2 * 3 = 6 and 2 + 3 = 5)

Set each factor to zero: (x + 2) = 0 or (x + 3) = 0 Solve: x = -2 or x = -3 Example 2: Solve 2x² - x - 3 = 0 The equation is already in standard form.

Factorize: 2x² - x - 3 = (2x - 3)(x + 1) (This can be found through trial and error or using techniques for factoring when a is not 1)

Set each factor to zero: (2x - 3) = 0 or (x + 1) = 0 Solve: x = 3/2 or x = -1 2.1.2 Solving by Completing the Square: This method is useful when factorization is difficult or impossible.

Process: Rearrange the equation to the form ax² + bx = -c. Divide both sides by a (if a ≠ 1). Add (b/2a)² to both sides of the equation. This makes the left side a perfect square trinomial. Express the left side as a squared binomial. Take the square root of both sides (remembering both positive and negative roots). Solve for x.

Example: Solve x² + 4x - 1 = 0 x² + 4x = 1 a = 1, so no need to divide. Add (4/2)² = 4 to both sides: x² + 4x + 4 = 1 + 4 (x + 2)² = 5 x + 2 = ±√5 x = -2 ± √5 Therefore, x = -2 + √5 or x = -2 - √5 2.1.3 Solving using the Quadratic Formula: The quadratic formula provides a general solution for any quadratic equation. It is derived from the method of completing the square.

Formula: For ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a

Example: Solve 2x² + 5x - 3 = 0 Identify a = 2, b = 5, c = -3 Substitute into the formula: x = (-5 ± √(5² - 4 2 -3)) / (2 2)* Simplify: x = (-5 ± √(25 + 24)) / 4 = (-5 ± √49) / 4 = (-5 ± 7) / 4 Solve: x = (-5 + 7) / 4 = 2/4 = 1/2 or x = (-5 - 7) / 4 = -12/4 = -3 2.1.4 The Discriminant: The discriminant, Δ = b² - 4ac, provides information about the nature of the roots of the quadratic equation. Δ > 0: Two distinct real roots. Δ = 0: One real root (repeated root). Δ 0, ax² + bx + c 0 The inequality is already in the correct form. Solve x² - 3x + 2 = 0: (x - 1)(x - 2) = 0, so x = 1 or x =

2. These are the critical values.

Number line: ``` -----|-----|----- 1 2 ``` Test: x

0. Satisfies the inequality. 1 2: Let x = 3. (3)² - 3(3) + 2 = 2 >

0. Satisfies the inequality.

Solution: x

2. In interval notation: (-∞, 1) ∪ (2, ∞) 2.3 Sketching Quadratic Functions: The general form of a quadratic function is f(x) = ax² + bx + c.

Key Features: Shape: Parabola (U-shaped if a > 0, upside-down U-shaped if a

0. Does not satisfy. -1/2 3: Let x = 4. 2(4)² - 5(4) - 3 = 9 >

0. Does not satisfy. Since we have "≤", the solutions are included Solution: -1/2 ≤ x ≤

3. In interval notation: [-1/2, 3]* Question 5: Determine the nature of roots of the equation x² + 4x + 4 =

0. Solution: Find the discriminant: Δ = b² - 4ac = 4² - 4(1)(4) = 16 - 16 = 0 Since the discriminant is equal to 0, the quadratic equation has one real root. Independent Practice (Questions Only)

Solve: x² - 8x + 15 = 0 (by factorization)

Solve: x² + 2x - 8 = 0 (by completing the square)

Solve: x² - 3x - 5 = 0 (using the quadratic formula)

Solve: 4x² - 9 = 0 (by factorization)

Solve: x² - 4x > 0 Solve: x² + 2x - 3 < 0 Solve: 2x² + x - 1 ≥ 0 Determine the nature of the roots of x² - 6x + 5 =

0. Determine the nature of the roots of x² + 2x + 3 =

0. Sketch the graph of y = x² - 4x +

3. Indicate the turning point and intercepts with the axes.