Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Quadratic Equation
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Quadratic Equation
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Subject: General Mathematics
Class: Senior Secondary 1
Term: 1st Term
Week: 2
Theme: Algebraic Process
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Solve in volving problems factorization of quadratic expressions Solve quadratic equation of the form ab =0 a =0 or b =0 Form quadratic equation with given roots Draw quadratic equation graphs Read and obtain roots from a quadratic graph Solve word problems in volving real life situations
Algebraic Process Quadratic Equation Term: 1st Term Week: 6 ---
1. Overview and Learning Objectives This topic introduces students to quadratic equations, a fundamental concept in algebra with wide-ranging applications in science, engineering, economics, and everyday problem-solving. Unlike linear equations which have a single solution, quadratic equations typically have two solutions, or roots, reflecting the parabolic nature of their graphs. Understanding quadratic equations equips students with crucial analytical skills necessary for advanced mathematical studies and practical real-world scenarios in Nigeria, such as calculating the trajectory of objects, designing architectural structures, or optimizing economic models.
Specific Learning Objectives: By the end of this lesson, students will be able to:
1. Solve problems involving the factorization of quadratic expressions.
2. Solve quadratic equations of the form $ab = 0$, where $a=0$ or $b=0$.
3. Form quadratic equations when given their roots.
4. Draw graphs of quadratic equations.
5. Read and obtain the roots of a quadratic equation from its graph.
6. Solve word problems involving real-life situations by translating them into quadratic equations.
Real-world Connections in Nigeria: Factorization: Used in determining optimal dimensions for rectangular plots of land for farming or building in Nigeria, where area or perimeter constraints exist.
Forming Equations from Roots: Applicable in engineering for designing structures with specific stress points, or in economics for modeling scenarios with known break-even points.
Quadratic Graphs: Essential for understanding projectile motion in sports (e.g., a football kick, javelin throw) or for optimizing the path of water from an irrigation sprinkler in a Nigerian farm. Also used in architectural design of parabolic arches for bridges or buildings.
Word Problems: Used to solve practical issues such as calculating the number of livestock given certain conditions, determining the speed of a vehicle, or optimizing profit margins for a small business selling local produce or crafts.
2. Key Concepts and Explanations 2.1 Definition of a Quadratic Equation A quadratic equation is a polynomial equation of the second degree.
Its general or standard form is given by: $ax^2 + bx + c = 0$ where: $x$ is the variable. $a$, $b$, and $c$ are real constants. $a \neq 0$ (If $a=0$, the equation becomes linear). $ax^2$ is the quadratic term. $bx$ is the linear term. $c$ is the constant term. The solutions to a quadratic equation are called its roots or zeros. These are the values of $x$ that make the equation true. 2.2 Methods of Solving Quadratic Equations 2.2.1 Solving by Factorization This method relies on the Zero Product Property: If the product of two or more factors is zero, then at least one of the factors must be zero. That is, if $AB = 0$, then $A=0$ or $B=0$.
Steps:
1. Ensure the quadratic equation is in the standard form $ax^2 + bx + c = 0$.
2. Find two numbers (let's call them $p$ and $q$) such that their product ($p \times q$) equals $ac$ and their sum ($p + q$) equals $b$.
3. Rewrite the middle term ($bx$) as the sum of $px$ and $qx$.
4. Factor by grouping the terms.
5. Apply the Zero Product Property to set each factor equal to zero and solve for $x$.
Worked Example 1 (Factorization - Simple): Solve $x^2 - 5x + 6 = 0$.
Solution:
1. The equation is already in standard form. ($a=1, b=-5, c=6$)
2. Find two numbers whose product is $ac = 1 \times 6 = 6$ and whose sum is $b = -5$. The numbers are $-2$ and $-3$ (since $(-2) \times (-3) = 6$ and $(-2) + (-3) = -5$).
3. Rewrite the middle term: $x^2 - 2x - 3x + 6 = 0$.
4. Factor by grouping: $x(x - 2) - 3(x - 2) = 0$ $(x - 2)(x - 3) = 0$
5. Apply Zero Product Property: $x - 2 = 0 \implies x = 2$ or $x - 3 = 0 \implies x = 3$ The roots are $x=2$ or $x=3$. Worked Example 2 (Factorization - with $a \neq 1$): A rectangular plot of land in Abuja has an area of 42 square meters. Its length is 1 meter more than are $x=5/2$ or $x=-1$. 2.3 Forming Quadratic Equations with Given Roots There are two main methods: Method 1: Using the Zero Product Property (Factor Method) If $\alpha$ and $\beta$ are the roots of a quadratic equation, then $(x - \alpha)$ and $(x - \beta)$ are its factors.
Therefore, the equation is $(x - \alpha)(x - \beta) = 0$. Expanding this gives $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Method 2: Using the Sum and Product of Roots For a quadratic equation $ax^2 + bx + c = 0$, dividing by $a$ gives $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$. If $\alpha$ and $\beta$ are the roots, then: Sum of roots: $\alpha + \beta = -\frac{b}{a}$ Product of roots: $\alpha\beta = \frac{c}{a}$ So, the equation can be written as: $x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$ $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ Worked Example 5 (Forming Quadratic Equations): Form a quadratic equation whose roots are $1/2$ and $-3$.
Solution (Method 1 - Factor Method):
1. Let $\alpha = 1/2$ and $\beta = -3$.
2. The factors are $(x - 1/2)$ and $(x - (-3))$, which is $(x - 1/2)$ and $(x + 3)$.
3. Set their product to zero: $(x - 1/2)(x + 3) = 0$
4. Expand the expression: $x(x + 3) - 1/2(x + 3) = 0$ $x^2 + 3x - 1/2x - 3/2 = 0$ $x^2 + (6/2 - 1/2)x - 3/2 = 0$ $x^2 + 5/2x - 3/2 = 0$
5. To clear fractions, multiply the entire equation by 2: $2(x^2 + 5/2x - 3/2) = 2(0)$ $2x^2 + 5x - 3 = 0$ Solution (Method 2 - Sum and Product of Roots):
1. Let $\alpha = 1/2$ and $\beta = -3$.
2. Calculate the sum of roots: $\alpha + \beta = 1/2 + (-3) = 1/2 - 3 = 1/2 - 6/2 = -5/2$
3. Calculate the product of roots: $\alpha\beta = (1/2) \times (-3) = -3/2$
4. Substitute into the formula $x^2 - (\alpha + \beta)x + \alpha\beta = 0$: $x^2 - (-5/2)x + (-3/2) = 0$ $x^2 + 5/2x - 3/2 = 0$
5. Multiply by 2 to clear fractions: $2x^2 + 5x - 3 = 0$ Both methods yield the same quadratic equation. 2.4 Quadratic Equation Graphs (Parabolas) The graph of a quadratic equation $y = ax^2 + bx + c$ is a parabola.
Key Features: Shape: If $a > 0$, the parabola opens upwards (U-shape), and its vertex is a minimum point. If $a 0$ or $< 0$) and represent the solution sets graphically and algebraically.
Transformations of Quadratic Graphs: Explore how changing the values of $a, b, c$ in $y = ax^2 + bx + c$ affects the position, shape, and orientation of the parabola (e.g., vertex form $y=a(x-h)^2+k$).
Real-world Optimization Problems: Provide more complex word problems that require optimization techniques beyond simple area/perimeter, perhaps involving profit maximization or cost minimization scenarios with more variables.
Cubic Equations: Briefly introduce higher-order polynomial equations and methods for solving them (e.g., factor theorem), showing the progression beyond quadratic. Since width cannot be negative, we reject $w = -3$.
1
0. The width is $w = 2.5$ meters.
1
1. The length is $l = 2w + 1 = 2(2.5) + 1 = 5 + 1 = 6$ meters. The dimensions of the poultry pen are $2.5 \text{ m}$ by $6 \text{ m}$. 2.2.2 Solving by Completing the Square This method converts the quadratic expression into a perfect square trinomial. This is crucial for deriving the quadratic formula.
Steps:
1. Ensure the equation is in the form $ax^2 + bx + c = 0$.
2. If $a \neq 1$, divide the entire equation by $a$.
3. Move the constant term to the right side of the equation.
4. Take half of the coefficient of the $x$ term (which is $b/a$ from previous step), square it $(\frac{b}{2a})^2$, and add it to both sides of the equation.
5. Factor the left side as a perfect square $(x + \frac{b}{2a})^2$.
6. Take the square root of both sides, remembering the $\pm$ sign.
7. Solve for $x$.
Worked Example 3 (Completing the Square): Solve $x^2 + 6x + 5 = 0$ using the method of completing the square.
Solution: 1. $a=1$, so no division needed.
2. Move constant term: $x^2 + 6x = -5$.
3. Half of the coefficient of $x$ is $6/2 = 3$.
Square it: $3^2 = 9$.
Add 9 to both sides: $x^2 + 6x + 9 = -5 + 9$
4. Factor the left side as a perfect square: $(x + 3)^2 = 4$
5. Take the square root of both sides: $x + 3 = \pm\sqrt{4}$ $x + 3 = \pm2$
6. Solve for $x$: $x + 3 = 2 \implies x = 2 - 3 = -1$ or $x + 3 = -2 \implies x = -2 - 3 = -5$ The roots are $x=-1$ or $x=-5$. 2.2.3 Solving using the Quadratic Formula The quadratic formula is a general solution for any quadratic equation in standard form.
Derivation of the Quadratic Formula: Start with the general form: $ax^2 + bx + c = 0$
1. Divide by $a$ (assuming $a \neq 0$): $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Move constant term to the right: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Complete the square on the left. Add $(\frac{1}{2} \cdot \frac{b}{a})^2 = (\frac{b}{2a})^2 = \frac{b^2}{4a^2}$ to both sides: $x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$
4. Factor the left side and simplify the right side: $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$
5. Take the square root of both sides: $x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$ $x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$
6. Isolate $x$: $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ This is the quadratic formula.
Steps for Application:
1. Ensure the equation is in standard form $ax^2 + bx + c = 0$.
2. Identify the values of $a$, $b$, and $c$.
3. Substitute these values into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
4. Simplify to find the roots.
Worked Example 4 (Quadratic Formula): Solve $2x^2 - 3x - 5 = 0$ using the quadratic formula.
Solution:
1. The equation is in standard form.
2. Identify $a=2, b=-3, c=-5$.
3. Substitute into the formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$ $x = \frac{3 \pm \sqrt{9 - (-40)}}{4}$ $x = \frac{3 \pm \sqrt{9 + 40}}{4}$ $x = \frac{3 \pm \sqrt{49}}{4}$ $x = \frac{3 \pm 7}{4}$
4. Solve for the two roots: $x_1 = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$ $x_2 = \frac{3 - 7}{4} = \frac{-4}{4} = -1$ The roots are $x=5/2$ or $x=-1$. 2.3 Forming Quadratic Equations with Given Roots There are two main methods: Method 1: Using the Zero Product Property (Factor Method) If $\alpha$ and $\beta$ are the roots of a quadratic equation, then $(x - \alpha)$ and $(x - \beta)$ are its factors.
Therefore, the equation is $(x - \alpha)(x - \beta) = 0$. Expanding this gives $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Method 2: Using the Sum and Product of Roots For a quadratic equation $ax^2 + bx + c = 0$, dividing by