### Quadratic Equation #### Learning Objectives: 1. Define a quadratic equation. 2. Identify different methods of solving quadratic equations. 3. Solve quadratic equations by factorization. 4. Solve quadratic equations by completing the square. 5. Solve quadratic equations using the general formula. #### 1. Definition of a Quadratic Equation A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term in which the unknown variable is squared. Its standard form is $ax^2 + bx + c = 0$, where $x$ is the variable, and $a$, $b$, and $c$ are real numbers with $a \neq 0$. The roots or solutions of a quadratic equation are the values of $x$ that satisfy the equation. #### 2. Methods of Solving Quadratic Equations There are several methods to solve quadratic equations: * **Factorization** * **Completing the Square** * **Using the General Formula (Quadratic Formula)** * Graphical Method (not covered in detail today) #### 3. Solving by Factorization This method involves rewriting the quadratic expression as a product of two linear factors. If $ (x-p)(x-q) = 0 $, then $x-p=0$ or $x-q=0$, which means $x=p$ or $x=q$. This method is suitable when the quadratic expression can be easily factorized. **Worked Examples:** **EASY**: Solve $x^2 - 5x + 6 = 0$ * Find two numbers that multiply to 6 and add to -5. These are -2 and -3. * Rewrite the equation: $(x-2)(x-3) = 0$ * Set each factor to zero: $x-2=0$ or $x-3=0$ * Solutions: $x=2$ or $x=3$ **MEDIUM**: Solve $2x^2 + 7x + 3 = 0$ * Multiply 'a' and 'c': $2 \times 3 = 6$. Find two numbers that multiply to 6 and add to 7. These are 1 and 6. * Rewrite the middle term: $2x^2 + x + 6x + 3 = 0$ * Factor by grouping: $x(2x+1) + 3(2x+1) = 0$ * Factor out the common bracket: $(2x+1)(x+3) = 0$ * Set each factor to zero: $2x+1=0$ or $x+3=0$ * Solutions: $x = -\frac{1}{2}$ or $x = -3$ **EXAM**: Solve $6x^2 - x - 12 = 0$ * Multiply 'a' and 'c': $6 \times -12 = -72$. Find two numbers that multiply to -72 and add to -1. These are -9 and 8. * Rewrite the middle term: $6x^2 - 9x + 8x - 12 = 0$ * Factor by grouping: $3x(2x-3) + 4(2x-3) = 0$ * Factor out the common bracket: $(2x-3)(3x+4) = 0$ * Set each factor to zero: $2x-3=0$ or $3x+4=0$ * Solutions: $x = \frac{3}{2}$ or $x = -\frac{4}{3}$ #### 4. Solving by Completing the Square This method involves manipulating the quadratic equation into the form $ (x+p)^2 = q $, then taking the square root of both sides. It is particularly useful when factorization is difficult. **Steps:** 1. Divide the entire equation by $a$ if $a \neq 1$. 2. Move the constant term ($c/a$) to the right side of the equation. 3. Take half of the coefficient of $x$ (i.e., $(b/2a)$), square it ($(b/2a)^2$), and add it to both sides of the equation. 4. Factor the left side as a perfect square trinomial. 5. Take the square root of both sides (remembering $\pm$). 6. Solve for $x$. **Worked Examples:** **MEDIUM**: Solve $x^2 + 6x - 7 = 0$ * Move constant term: $x^2 + 6x = 7$ * Half of coefficient of $x$ is $(6/2)=3$. Square it: $3^2=9$. Add to both sides: $x^2 + 6x + 9 = 7 + 9$ * Factor left side: $(x+3)^2 = 16$ * Take square root: $x+3 = \pm\sqrt{16}$ * $x+3 = \pm 4$ * Solutions: $x = -3 + 4 = 1$ or $x = -3 - 4 = -7$ **EXAM**: Solve $2x^2 - 4x - 6 = 0$ * Divide by $a=2$: $x^2 - 2x - 3 = 0$ * Move constant term: $x^2 - 2x = 3$ * Half of coefficient of $x$ is $(-2/2)=-1$. Square it: $(-1)^2=1$. Add to both sides: $x^2 - 2x + 1 = 3 + 1$ * Factor left side: $(x-1)^2 = 4$ * Take square root: $x-1 = \pm\sqrt{4}$ * $x-1 = \pm 2$ * Solutions: $x = 1 + 2 = 3$ or $x = 1 - 2 = -1$ #### 5. Solving using the General Formula (Quadratic Formula) The quadratic formula can be used to solve any quadratic equation, regardless of whether it is factorable or not. It is derived by completing the square on the general form $ax^2 + bx + c = 0$. The formula is: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ **Worked Examples:** **MEDIUM**: Solve $x^2 + 5x + 6 = 0$ * Identify $a=1, b=5, c=6$ * Substitute into formula: $x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(6)}}{2(1)}$ $x = \frac{-5 \pm \sqrt{25 - 24}}{2}$ $x = \frac{-5 \pm \sqrt{1}}{2}$ $x = \frac{-5 \pm 1}{2}$ * Solutions: $x = \frac{-5+1}{2} = \frac{-4}{2} = -2$ or $x = \frac{-5-1}{2} = \frac{-6}{2} = -3$ **EXAM**: Solve $3x^2 - 7x + 2 = 0$ * Identify $a=3, b=-7, c=2$ * Substitute into formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)}$ $x = \frac{7 \pm \sqrt{49 - 24}}{6}$ $x = \frac{7 \pm \sqrt{25}}{6}$ $x = \frac{7 \pm 5}{6}$ * Solutions: $x = \frac{7+5}{6} = \frac{12}{6} = 2$ or $x = \frac{7-5}{6} = \frac{2}{6} = \frac{1}{3}$ #### Board Summary: ``` +-------------------------------------------------------------+ | QUADRATIC EQUATION (SS 2) | +-------------------------------------------------------------+ | Definition: $ax^2 + bx + c = 0$, where $a \neq 0$ | | | | Methods: | | 1. Factorization: $(x-p)(x-q)=0 \implies x=p, x=q$ | | E.g., $x^2 - 5x + 6 = 0 \implies (x-2)(x-3)=0 \implies x=2,3$ | | | | 2. Completing the Square: Make one side a perfect square. | | E.g., $x^2 + 6x - 7 = 0 \implies (x+3)^2 = 16 \implies x=1,-7$ | | | | 3. General Formula: | | $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ | | E.g., $x^2 + 5x + 6 = 0 \implies x = \frac{-5 \pm 1}{2} \implies x=-2,-3$ | +-------------------------------------------------------------+ ``` #### Evaluation Questions: 1. What is the standard form of a quadratic equation? 2. Solve $x^2 - x - 12 = 0$ by factorization. 3. Solve $x^2 + 4x - 5 = 0$ by completing the square. 4. Solve $2x^2 + 3x - 2 = 0$ using the general formula. #### Assignment Questions: 1. Solve the following quadratic equations by any method: a) $x^2 - 7x + 10 = 0$ b) $3x^2 + 5x - 2 = 0$ c) $x^2 - 8x + 15 = 0$ d) $5x^2 + 13x + 6 = 0$ e) $x^2 - 2x - 1 = 0$ (use completing the square or formula)