Word Problems Involving Quadratic Equations Notes

<definition term="Quadratic Equation">An equation of the form \\$ax^2 + bx + c = 0\\$, where \\$a, b, c\\$ are constants and \\$a \neq 0\\$.</definition> <callout type="note">Quadratic equations are **non-linear** equations, meaning that their graphs are not straight lines. </callout> ### Solving Quadratic Equations There are three main methods to solve quadratic equations: 1. **Factoring** 2. **Completing the square** 3. **Quadratic formula** <callout type="tip">When solving quadratic equations, always start by **simplifying** the equation as much as possible. </callout> #### Solving by Factoring <definition term="Factoring">The process of expressing a polynomial as a product of its factors.</definition> To solve a quadratic equation by factoring: 1. **Rewrite** the equation in the form \\$ax^2 + bx + c = 0\\$. 2. **Factor** the quadratic expression on the left side. 3. **Set** each factor equal to zero and solve for the variable. <callout type="example">Solve the equation \\$x^2 - 2x - 3 = 0\\$ by factoring. 1. The equation is already in the form \\$ax^2 + bx + c = 0\\$. 2. Factor the quadratic expression: \\$x^2 - 2x - 3 = (x + 1)(x - 3)\\$. 3. Set each factor equal to zero: * \\$x + 1 = 0 \implies x = -1\\$ * \\$x - 3 = 0 \implies x = 3\\$ The solutions are \\$x = -1\\$ and \\$x = 3\\$. </callout> <callout type="note">Not all quadratic equations can be factored easily. In such cases, use the quadratic formula or completing the square. </callout> #### Solving by Completing the Square <definition term="Completing the Square">A method of solving quadratic equations by transforming the equation into a perfect square trinomial.</definition> To solve a quadratic equation by completing the square: 1. **Rewrite** the equation in the form \\$ax^2 + bx = -c\\$. 2. **Divide** both sides by \\$a\\$ if \\$a \neq 1\\$. 3. **Add** \\$(\frac{b}{2})^2\\$ to both sides to complete the square. 4. **Rewrite** the left side as a perfect square trinomial. 5. **Solve** for the variable by taking the square root of both sides. <callout type="example">Solve the equation \\$x^2 - 2x - 3 = 0\\$ by completing the square. 1. Rewrite the equation: \\$x^2 - 2x = 3\\$. 2. Add \\$(\frac{-2}{2})^2 = 1\\$ to both sides: \\$x^2 - 2x + 1 = 4\\$. 3. Rewrite the left side as a perfect square trinomial: \\$(x - 1)^2 = 4\\$. 4. Take the square root of both sides: \\$x - 1 = \pm 2\\$. 5. Solve for \\$x\\$: \\$x = 1 + 2 = 3\\$ or \\$x = 1 - 2 = -1\\$. The solutions are \\$x = -1\\$ and \\$x = 3\\$. </callout> #### Solving by the Quadratic Formula <definition term="Quadratic Formula">A formula that provides the solutions to a quadratic equation \\$ax^2 + bx + c = 0\\$: \\$\\$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\$\\$</definition> To solve a quadratic equation using the quadratic formula: 1. **Identify** the coefficients \\$a\\$, \\$b\\$, and \\$c\\$. 2. **Substitute** the values into the quadratic formula. 3. **Simplify** to find the solutions. <callout type="example">Solve the equation \\$x^2 - 2x - 3 = 0\\$ using the quadratic formula. 1. Identify the coefficients: \\$a = 1\\$, \\$b = -2\\$, \\$c = -3\\$. 2. Substitute into the quadratic formula: \\$\\$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}\\$\\$ 3. Simplify: \\$\\$x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}\\$\\$ 4. Solve for \\$x\\$: \\$x = \frac{6}{2} = 3\\$ or \\$x = \frac{-2}{2} = -1\\$. The solutions are \\$x = -1\\$ and \\$x = 3\\$. </callout> <callout type="note">The **discriminant** \\$b^2 - 4ac\\$ determines the nature of the solutions: * If \\$b^2 - 4ac > 0\\$, there are two distinct real solutions. * If \\$b^2 - 4ac = 0\\$, there is one repeated real solution. * If \\$b^2 - 4ac < 0\\$, there are no real solutions (the solutions are complex). </callout> ### Graphing Quadratic Equations The graph of a quadratic equation \\$y = ax^2 + bx + c\\$ is a **parabola**. <callout type="note">The **vertex** of the parabola is the point \\$(h, k)\\$ where \\$h = -\frac{b}{2a}\\$ and \\$k = f(h)\\$. </callout> #### Graphing by Hand To graph a quadratic equation by hand: 1. **Find** the vertex \\$(h, k)\\$. 2. **Determine** the direction of the parabola (upward if \\$a > 0\\$, downward if \\$a < 0\\$). 3. **Find** the \\$y\\$-intercept \\$(0, c)\\$. 4. **Find** the \\$x\\$-intercepts by solving the equation \\$ax^2 + bx + c = 0\\$. 5. **Plot** additional points if needed and draw the parabola. <callout type="example">Graph the equation \\$y = x^2 - 2x - 3\\$. 1. Find the vertex: \\$h = -\frac{-2}{2(1)} = 1\\$, \\$k = (1)^2 - 2(1) - 3 = -4\\$. The vertex is \\$(1, -4)\\$. 2. The parabola opens upward since \\$a = 1 > 0\\$. 3. The \\$y\\$-intercept is \\$(0, -3)\\$. 4. The \\$x\\$-intercepts are \\$(-1, 0)\\$ and \\$(3, 0)\\$ (from the solutions of \\$x^2 - 2x - 3 = 0\\$). 5. Plot the points and draw the parabola. <placeholder type="graph" description="Graph of the parabola y = x^2 - 2x - 3 with vertex at (1, -4), y-intercept at (0, -3), and x-intercepts at (-1, 0) and (3, 0)."></placeholder> </callout> #### Graphing with a Calculator To graph a quadratic equation using a calculator: 1. **Enter** the equation into the calculator's graphing function. 2. **Adjust** the window settings to view the relevant portion of the graph. 3. **Use** the calculator's features to find the vertex, intercepts, and other key points. <callout type="note">Graphing calculators are useful for checking your work and visualizing the behavior of quadratic functions. </callout> ### Focus and Directrix of a Parabola <definition term="Focus">A point inside the parabola that is equidistant from the directrix and any point on the parabola.</definition> <definition term="Directrix">A line outside the parabola that is equidistant from the focus and any point on the parabola.</definition> The focus and directrix provide an alternative way to define a parabola. <callout type="note">The distance between the vertex and the focus (or directrix) is denoted by \\$p\\$. </callout> #### Finding the Focus and Directrix For a parabola in the form \\$y = ax^2 + bx + c\\$: 1. **Find** the vertex \\$(h, k)\\$. 2. **Calculate** \\$p = \frac{1}{4a}\\$. 3. The focus is at \\$(h, k + p)\\$ and the directrix is at \\$y = k - p\\$. <callout type="example">Find the focus and directrix of the parabola \\$y = x^2 - 2x - 3\\$. 1. The vertex is \\$(1, -4)\\$ (from previous example). 2. Calculate \\$p = \frac{1}{4(1)} = \frac{1}{4}\\$. 3. The focus is at \\$(1, -4 + \frac{1}{4}) = (1, -\frac{15}{4})\\$. 4. The directrix is at \\$y = -4 - \frac{1}{4} = -\frac{17}{4}\\$. <placeholder type="graph" description="Graph of the parabola y = x^2 - 2x - 3 with focus at (1, -15/4) and directrix at y = -17/4."></placeholder> </callout> <callout type="self_review">1. Solve the equation \\$2x^2 - 3x - 5 = 0\\$ using all three methods. 2. Graph the equation \\$y = -x^2 + 4x - 3\\$ and find the vertex, intercepts, focus, and directrix. </callout> <callout type="tok">How do different methods of solving quadratic equations reflect the nature of mathematical problem-solving? What are the advantages and limitations of each method? </callout> <callout type="case_study">**Application of Quadratic Equations in Physics:** Quadratic equations are used to model the motion of objects under constant acceleration, such as projectiles. The equation \\$s = ut + \frac{1}{2}at^2\\$ is a quadratic equation in \\$t\\$, where \\$s\\$ is the displacement, \\$u\\$ is the initial velocity, \\$a\\$ is the acceleration, and \\$t\\$ is the time. By solving this equation, we can determine the time it takes for an object to reach a certain height or return to the ground. </callout>