Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Solve quadratic equations using factorization, completing the square, and the quadratic formula.
• Interpret the significance of non-real roots.
• Solve quadratic inequalities.
• Apply this knowledge to solve real-world problems.

Lesson summary

This week, we're diving deeper into the world of equations and inequalities, building upon the foundational knowledge you acquired in earlier grades. Understanding equations and inequalities is crucial not just for mathematics but also for making informed decisions in your everyday life. Whether you're budgeting your money, comparing cell phone contracts, or analyzing data, the principles we'll cover this week are essential tools. In South Africa, where resource management and financial literacy are vital, mastering these concepts empowers you to become more responsible and informed citizens.

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 a quadratic equation are called its roots or zeros. 2.1.1 Solving by Factorization This method works when the quadratic expression ax² + bx + c can be factored into two linear factors.

Example 1: Solve x² + 5x + 6 =

0. Step 1: Factor the quadratic expression: (x + 2)(x + 3) = 0 Step 2: Set each factor equal to zero: x + 2 = 0 or x + 3 = 0 Step 3: Solve for x: x = -2 or x = -3 Therefore, the roots are x = -2 and x = -

3. Example 2: Solve 2x² - 7x + 3 =

0. Step 1: Factor the quadratic expression: (2x - 1)(x - 3) = 0 Step 2: Set each factor equal to zero: 2x - 1 = 0 or x - 3 = 0 Step 3: Solve for x: x = 1/2 or x = 3 Therefore, the roots are x = 1/2 and x = 3. 2.1.2 Solving by Completing the Square This method is useful for solving any quadratic equation, especially when factorization is difficult.

Example 3: Solve x² + 6x + 5 = 0 by completing the square.

Step 1: Move the constant term to the right side: x² + 6x = -5 Step 2: Take half of the coefficient of the x term (which is 6), square it (3² = 9), and add it to both sides: x² + 6x + 9 = -5 + 9 Step 3: Factor the left side as a perfect square: (x + 3)² = 4 Step 4: Take the square root of both sides: x + 3 = ±2 Step 5: Solve for x: x = -3 ± 2 This gives us x = -1 or x = -

5. Therefore, the roots are x = -1 and x = -5. 2.1.3 Solving by the Quadratic Formula The quadratic formula is a general formula for solving any quadratic equation ax² + bx + c = 0: x = (-b ± √(b² - 4ac)) / 2a Example 4: Solve 3x² - 5x + 2 = 0 using the quadratic formula.

Step 1: Identify a, b, and c: a = 3, b = -5, c = 2 Step 2: Substitute the values into the quadratic formula: x = (5 ± √((-5)² - 4 3 2)) / (2 3)* x = (5 ± √(25 - 24)) / 6 x = (5 ± √1) / 6 x = (5 ± 1) / 6 Step 3: Solve for x: x = (5 + 1) / 6 = 1 or x = (5 - 1) / 6 = 2/3 Therefore, the roots are x = 1 and x = 2/3. 2.1.4 Non-Real Roots The discriminant, Δ = b² - 4ac, determines the nature of the roots. If Δ > 0, the equation has two distinct real roots. If Δ = 0, the equation has one real root (a repeated root). If Δ 0, ax² + bx + c

0. Step 1: Find the roots of the corresponding quadratic equation x² - 3x - 4 =

0. Factoring gives us (x - 4)(x + 1) = 0, so x = 4 or x = -

1. These are critical values.

Step 2: Draw a number line and mark the critical values (-1 and 4) on it. These values divide the number line into three intervals: x

4. Step 3: Choose a test value from each interval and substitute it into the inequality to determine if the inequality is true or false in that interval. For x

0. The inequality is true. For -1

0. The inequality is false. For x > 4, let x = 5: (5)² - 3(5) - 4 = 25 - 15 - 4 = 6 >

0. The inequality is true.

Step 4: Write the solution set. Since we want the values where x² - 3x - 4 > 0, the solution is x

4. In interval notation, the solution is (-∞, -1) ∪ (4, ∞).

Example 7: Solve -x² + 5x - 6 ≤

0. Step 1: Multiply by -1 (remember to reverse the inequality): x² - 5x + 6 ≥

0. Step 2: Find the roots of x² - 5x + 6 =

0. Factoring gives (x - 2)(x - 3) = 0, so x = 2 or x =

3. Step 3: Test intervals: x

3. For x 3, let x = 4: (4)² - 5(4) + 6 = 16 - 20 + 6 = 2 ≥

0. True.

Step 4: Write the solution set, including the roots since it's ≥ 0: x ≤ 2 or x ≥

3. In interval notation, the solution is (-∞, 2] ∪ [3, ∞). Guided Practice (With Solutions)

Question 1: Solve the equation x² - 8x + 15 = 0 by factorization.

Solution: Step 1: Factor the quadratic expression: (x - 3)(x - 5) = 0 Step 2: Set each factor equal to zero: x - 3 = 0 or x - 5 = 0 Step 3: Solve for x: x = 3 or x = 5 Therefore, the roots are x = 3 and x =

5. Commentary: This question tests the basic skill of solving quadratic equations by factorization. We identify two numbers that add up to -8 and multiply to 15 (-3 and -5).

Question 2: Solve the equation 2x² + 4x - 3 = 0 using the quadratic formula.

Solution: Step 1: Identify a, b, and c: a = 2, b = 4, c = -3 Step 2: Substitute the values into the quadratic formula: x = (-4 ± √((4)² - 4 2 -3)) / (2 2)* x = (-4 ± √(16 + 24)) / 4 x = (-4 ± √40) / 4 x = (-4 ± 2√10) / 4 x = (-2 ± √10) / 2 Therefore, the roots are x = (-2 + √10) / 2 and x = (-2 - √10) /

2. Commentary: This question assesses the ability to correctly apply the quadratic formula. Simplifying the radical is also an important step.

Question 3: Solve the inequality x² + 2x - 8

2. For x 2, let x = 3: (3)² + 2(3) - 8 = 9 + 6 - 8 = 7

0. Solve the inequality: -2x² + 8x - 6 <

0. Solve the inequality: x² + 4x + 4 ≥

0. Determine the nature of the roots for the equation: 2x² - 3x + 5 =

0. Determine the values of k for which the equation x² + kx + 9 = 0 has real and equal roots.