A quadratic equation is a type of polynomial equation of the second degree, which means its highest exponent is two. In its most general form, it can be written as $ ax^2 + bx + c = 0 $, where $ a $, $ b $, and $ c $ are constants with $ a \neq 0 $. The solutions to a quadratic equation are the values of $ x $ that make the equation true, and these solutions are also known as the roots of the equation.
The standard form of a quadratic equation is an algebraic expression of the second degree in a variable, typically written as:
$ ax^2 + bx + c = 0, $
where:
In this form, $ ax^2 $ is the quadratic term, $ bx $ is the linear term, and $ c $ is the constant term.
The standard form is important because it reveals the coefficients that are used in various methods for solving the quadratic equation, such as factoring, completing the square, and applying the quadratic formula. Additionally, it's the most recognizable form that can be easily converted to other forms such as the vertex form or the factored form if needed for analysis or graphing.
To illustrate:
$ y = a(x - h)^2 + k, $
where $ (h, k) $ is the vertex of the parabola.
$ y = a(x - r_1)(x - r_2), $
where $ r_1 $ and $ r_2 $ are the solutions to the quadratic equation $ ax^2 + bx + c = 0 $.
When dealing with a quadratic equation, it is crucial that $ a $ does not equal zero. If $ a $ were zero, the equation would be linear, not quadratic, because the $ x^2 $ term would be eliminated, leaving only the $ bx + c $ terms, which represent a first-degree polynomial.
Each coefficient in the standard form of a quadratic equation gives information about the parabola:
Understanding the standard form of a quadratic equation is key to solving quadratic problems, analyzing parabolas graphically, and applying these concepts across various fields where quadratic relationships occur.
The quadratic formula is a solution method for quadratic equations that provides the roots (solutions) directly. It's derived from the process of completing the square on the general quadratic equation. Here's the formula:
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Let's break down its derivation and how to use it:
Start with the standard form of a quadratic equation:
$ ax^2 + bx + c = 0 $
$ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 $
$ x^2 + \frac{b}{a}x = -\frac{c}{a} $
$ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 $
Notice that the left side is now a perfect square trinomial, which can be factored into:
$ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} $
Combine the terms on the right side under a common denominator:
$ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} $
$ x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} $
$ x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} $
Simplify to get the quadratic formula:
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Solve the quadratic equation $ 2x^2 - 4x - 6 = 0 $ using the quadratic formula.
$ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)} $
$ x = \frac{4 \pm \sqrt{16 + 48}}{4} $
$ x = \frac{4 \pm \sqrt{64}}{4} $
$ x = \frac{4 \pm 8}{4} $
$ x = \frac{4 + 8}{4} = 3 $ $ x = \frac{4 - 8}{4} = -1 $
Answer: The solutions to the quadratic equation $ 2x^2 - 4x - 6 = 0 $ are $ x = 3 $ and $ x = -1 $.
The quadratic formula is a robust tool because it works for any quadratic equation, regardless of whether it is factorable by simple methods. It is especially helpful when the equation has irrational or complex roots, which cannot be easily factored.
Problem: Solve the quadratic equation $ x^2 - 7x + 10 = 0 $ .
Solution:
Answer: The solutions are $ x = 5 $ and $ x = 2 $.
Problem: Solve the quadratic equation $ x^2 + 6x - 7 = 0 $ by completing the square.
Solution:
Answer: The solutions are $ x = 1 $ and $ x = -7 $.
Problem: Use the quadratic formula to solve the equation $ 3x^2 - 2x - 1 = 0 $.
Solution:
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)} $
$ x = \frac{2 \pm \sqrt{4 + 12}}{6} $
$ x = \frac{2 \pm \sqrt{16}}{6} $
$ x = \frac{2 \pm 4}{6} $
Answer: The solutions are $ x = 1 $ and $ x = -\frac{1}{3} $.
Solving quadratic equations can be done through several methods, each with its own set of procedures. Here are the most common methods used to solve quadratic equations:
Factoring involves expressing the quadratic equation as the product of two binomial expressions (if possible). This method works well when the quadratic equation is factorable over the integers.
Steps for factoring:
Example: Solve $ x^2 - 5x + 6 = 0 $ by factoring. The factors of $ +6 $ that add up to $ -5 $ are $ -2 $ and $ -3 $:
$ (x - 2)(x - 3) = 0 $
Set each factor equal to zero and solve for $ x $:
$ x - 2 = 0 \quad \text{or} \quad x - 3 = 0 $
$ x = 2 \quad \text{or} \quad x = 3 $
Completing the square involves creating a perfect square trinomial from the quadratic equation, which can then easily be solved.
Steps for completing the square:
Example: Solve $ 2x^2 - 8x + 6 = 0 $ by completing the square. First, divide by 2:
$ x^2 - 4x + 3 = 0 $
Move the 3 to the other side:
$ x^2 - 4x = -3 $
Add $ (4/2)^2 = 4 $ to both sides:
$ x^2 - 4x + 4 = 1 $
Factor the left side:
$ (x - 2)^2 = 1 $
Take the square root of both sides:
$ x - 2 = \pm 1 $
$ x = 2 \pm 1 $
$ x = 3 \quad \text{or} \quad x = 1 $
The quadratic formula is the most general method that can solve any quadratic equation.
Quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Steps for using the quadratic formula:
Example: Solve $ x^2 + 6x - 7 = 0 $ using the quadratic formula. Here, $ a = 1 $, $ b = 6 $, and $ c = -7 $. Substitute these into the formula:
$ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-7)}}{2(1)} $
$ x = \frac{-6 \pm \sqrt{36 + 28}}{2} $
$ x = \frac{-6 \pm \sqrt{64}}{2} $
$ x = \frac{-6 \pm 8}{2} $
$ x = -7 \quad \text{or} \quad x = 1 $
Graphing involves plotting the quadratic equation on a coordinate plane and identifying the points where the parabola crosses the x-axis, which represent the solutions of the equation.
Steps for graphing:
Each of these methods has its own advantages, and the choice of method can depend on the specific form of the quadratic equation and the context in which the equation is being solved.
The range of a quadratic function refers to the set of all possible output values (y-values) it can produce. The range of the function depends on the orientation of the parabola that the quadratic equation represents when graphed on the coordinate plane.
Let's consider the standard form of a quadratic function:
$ f(x) = ax^2 + bx + c, $
where $ a $, $ b $, and $ c $ are constants, and $ a \neq 0 $.
Here's how to determine the range of a quadratic function:
Identify the Orientation of the Parabola:
Find the Vertex: The vertex of the parabola is the point $ (h, k) $, which represents the minimum or maximum value of the quadratic function. To find the vertex, use the formulas:
Determine the Range:
Consider the quadratic function:
$ f(x) = 2x^2 - 4x + 1. $
Since $ a = 2 $ is positive, the parabola opens upward.
To find the vertex:
With the parabola opening upward and the vertex being $ (1, -1) $, the range of the function is $ [-1, \infty) $.
Therefore, for this quadratic function, the output values $ y $ can be any real number that is $ -1 $ or greater.
Question: A ball is thrown upward from the top of a 20-meter-high building with an initial velocity of 15 m/s. How long will it take for the ball to hit the ground? Assume the acceleration due to gravity is $ -9.8 , \text{m/s}^2 $ (negative sign indicates downward direction).
Solution: The height $ h(t) $ of the ball at time $ t $ can be modeled by a quadratic equation derived from the kinematic equations:
$ h(t) = -\frac{1}{2} g t^2 + v_0 t + h_0, $
where $ g = 9.8 , \text{m/s}^2 $, $ v_0 = 15 , \text{m/s} $ is the initial velocity, and $ h_0 = 20 , \text{m} $ is the initial height.
Substituting the values, we get:
$ h(t) = -4.9t^2 + 15t + 20. $
To find out when the ball hits the ground, set $ h(t) $ to 0 and solve for $ t $:
$ 0 = -4.9t^2 + 15t + 20. $
Using the quadratic formula where $ a = -4.9 $, $ b = 15 $, and $ c = 20 $:
$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $
$ t = \frac{-15 \pm \sqrt{15^2 - 4(-4.9)(20)}}{2(-4.9)}. $
$ t = \frac{-15 \pm \sqrt{225 + 392}}{-9.8}. $
$ t = \frac{-15 \pm \sqrt{617}}{-9.8}. $
Since time cannot be negative, we take the positive root:
$ t = \frac{-15 + \sqrt{617}}{-9.8}. $
$ t \approx \frac{-15 + 24.86}{-9.8}. $
$ t \approx 1.007 \text{ seconds} $
So it will take approximately 1.007 seconds for the ball to reach the ground.
Question: A company manufactures and sells x items at a price of $ (200 - 0.5x) $ dollars per item. What is the number of items the company should sell to maximize its revenue?
Solution: Revenue $ R $ is the product of the number of items sold $ x $ and the price per item:
$ R(x) = x(200 - 0.5x). $
Expanding this, we get a quadratic equation:
$ R(x) = 200x - 0.5x^2. $
To maximize revenue, we need to find the vertex of the parabola represented by this quadratic function. The x-coordinate of the vertex gives the number of items for maximum revenue and can be found using $ -\frac{b}{2a} $ where $ a = -0.5 $ and $ b = 200 $:
$ x = -\frac{200}{2(-0.5)}. $
$ x = -\frac{200}{-1}. $
$ x = 200. $
Thus, the company should sell 200 items to maximize its revenue.
Question: A rectangular garden has a length that is 8 meters longer than its width. If the area of the garden is 240 square meters, what are the dimensions of the garden?
Solution: Let the width of the garden be $ w $ meters. Thus, the length will be $ w + 8 $ meters.
The area $ A $ of a rectangle is given by the product of its length and width:
$ A = \text{length} \times \text{width}. $
Substitute the given values:
$ 240 = w(w + 8). $
$ 240 = w^2 + 8w. $
Bring all terms to one side to form a standard quadratic equation:
$ w^2 + 8w - 240 = 0. $
This can be factored into:
$ (w - 12)(w + 20) = 0. $
Setting each factor equal to zero gives us the possible values for $ w $:
$ w - 12 = 0 \ \ \text{or} \ \ w + 20 = 0 $
$ w = 12 \ \ \text{or} \ \ w = -20. $
Since width cannot be negative, $ w = 12 $ meters.
The length is therefore:
$ w + 8 = 12 + 8 = 20 \text{ meters}. $
Hence, the dimensions of the garden are 12 meters in width and 20 meters in length.
Quadratic equations are essential in numerous fields because they can model many natural phenomena and are foundational in various scientific and applied disciplines. Here are several applications of quadratic equations:
In physics, the motion of a projectile thrown into the air, neglecting air resistance, can be modeled with a quadratic equation. The position of the projectile with respect to time is a parabola, with the equation given by:
$ y(t) = v_{0y}t - \frac{1}{2}gt^2 + h_0, $
where $ y(t) $ is the height at time $ t $, $ v_{0y} $ is the initial vertical velocity, $ g $ is the acceleration due to gravity, and $ h_0 $ is the initial height.
In engineering, quadratic equations can determine the load-bearing capacity of beams, the maximum stress an arch can withstand, or the shape of a cable suspended between two poles (a parabolic shape called a catenary).
In economics and business, quadratic equations help in optimization problems, such as maximizing profit or minimizing costs. For example, if a company's revenue or cost structure is quadratic with respect to the number of goods produced, it can use a quadratic equation to find the optimal production level.
The principles of optics often involve parabolas, which are described by quadratic equations. Lenses and mirrors with parabolic shapes focus light at a particular point, creating clearer images and more efficient energy concentration.
Ballistic trajectories, or the paths that bullets and rockets follow, are examples of parabolic motion. Quadratic equations are used to determine the trajectory of these projectiles, which is crucial for targeting and safety in military applications.
Quadratic equations can model population dynamics under certain conditions, such as the spread of a disease within a population or the growth rate of a species in an ecosystem when there are limiting factors.
In sports, quadratic equations can predict the path of balls in games like basketball, football, or golf. Understanding the projectile motion of the ball can help athletes and coaches make better decisions and improve performance.
Quadratic equations are sometimes involved in chemical kinetics, where they can represent the progression of reaction rates under specific conditions, particularly when dealing with second-order reactions.
Quadratic equations describe parabolic arches and other architectural elements that are not only aesthetically pleasing but also structurally efficient, distributing weight and resistance evenly.
In computer graphics, parabolic curves and surfaces, which can be represented by quadratic equations, are used to model objects and animations that appear in movies and video games, making them more lifelike.
In acoustics, the design of auditoriums and concert halls often uses parabolic surfaces to focus sound in specific directions, ensuring that audio is clear and evenly distributed throughout the space.
In number theory and cryptography, quadratic equations can be used in algorithms to find integers with particular properties or to solve problems related to prime numbers and factorization.
The applications mentioned are only a few examples, but they highlight the broad utility of quadratic equations in modeling and problem-solving in various practical situations. Understanding how to construct and solve these equations is valuable across many scientific and technical fields.
A quadratic equation is a second-degree polynomial equation in a single variable x with the general form:
$ ax^2 + bx + c = 0, $
where $ a $, $ b $, and $ c $ are constants with $ a \neq 0 $.
An equation is quadratic if it can be written in the form $ ax^2 + bx + c = 0 $ and $ a $ is not equal to zero. The highest power of the variable x is 2.
No, a quadratic equation in one variable has at most two distinct solutions. These solutions can be real or complex numbers.
If $ a $ is zero, the equation reduces to $ bx + c = 0 $, which is a linear equation, not a quadratic one. The defining feature of a quadratic equation is the $ ax^2 $ term.
The discriminant is the part of the quadratic formula under the square root, given by $ b^2 - 4ac $. It tells us the nature of the roots:
The vertex of a parabola represented by a quadratic equation is the point where the parabola turns. It is the maximum or minimum point on the graph, depending on whether the parabola opens upward or downward. The x-coordinate of the vertex can be found using the formula $ -b/(2a) $, and the y-coordinate can be found by substituting this x-value back into the equation.
Quadratic equations appear in various real-life situations, such as physics (projectile motion, determining the motion of objects under gravity), economics (profit maximization, cost functions), engineering (architecture and design), and even in sports (analyzing the flight paths of balls).
Completing the square involves rewriting a quadratic equation in the form $ (x + p)^2 = q $, where $ p $ and $ q $ are constants derived from the original coefficients. This creates a perfect square on one side of the equation, which can then be solved by taking the square root of both sides and isolating x.
The quadratic formula is a valuable tool because it provides a systematic way to find the solutions to any quadratic equation, regardless of whether the equation can be easily factored or not.