Quadratic Equations or Formula

What is a Quadratic Equation?

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 $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants with $a\ne 0$. 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.

Standard Form of Quadratic Equation

The standard form of a quadratic equation is an algebraic expression of the second degree in a variable, typically written as:

$a{x}^2+bx+c=0,$

where:

• $x$ is the variable or unknown that we are solving for,
• $a$, $b$, and $c$ are coefficients representing known numbers, and
• $a$ is non-zero.

In this form, $a{x}^2$ 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:

1. Vertex Form: This form of the quadratic function highlights the vertex (the highest or lowest point on the graph of the parabola) and is useful for graphing or analyzing the function's transformation. It can be written as:

$y=a(x-h{)}^2+k,$

where $(h,k)$ is the vertex of the parabola.

2. Factored Form: If a quadratic equation can be factored, its factored form shows the roots (or zeros) explicitly and is written as:

$y=a(x-r_1)(x-r_2),$

where $r_1$ and $r_2$ are the solutions to the quadratic equation $a{x}^2+bx+c=0$.

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:

• The sign of $a$ determines the direction the parabola opens. If $a$ is positive, it opens upward; if $a$ is negative, it opens downward.
• The value of $a$ affects the width and steepness of the parabola. Larger absolute values of $a$ make the parabola narrower, while smaller absolute values of $a$ make it wider.
• The $b$ coefficient, along with $a$, determines the location of the axis of symmetry of the parabola, which is the line that divides the parabola into two mirror images. The axis of symmetry can be found using the formula $x=-\frac{b}{2a}$.
• The $c$ coefficient represents the y-intercept of the parabola, which is the point where the graph crosses the y-axis.

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.

Quadratic Formula

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:

Derivation of the Quadratic Formula

Start with the standard form of a quadratic equation:

$a{x}^2+bx+c=0$

1. Divide all terms by $a$ (assuming $a$ is not zero):

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

2. Rearrange the equation so that the constant term is on the right side:

$x^2+\frac{b}{a}x=-\frac{c}{a}$

3. To complete the square, add the square of half the coefficient of $x$ to both sides of the equation:

$x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^2$

Notice that the left side is now a perfect square trinomial, which can be factored into:

${(x+\frac{b}{2a})}^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$

Combine the terms on the right side under a common denominator:

${(x+\frac{b}{2a})}^2=\frac{b^2-4ac}{4a^2}$

4. Take the square root of both sides, remembering to include both the positive and negative roots:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$

Simplify to get the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Components of the Quadratic Formula

• $x$: The variable representing the solutions to the quadratic equation.
• $a,b,c$: Coefficients from the quadratic equation $a{x}^2+bx+c=0$.
• $\pm$: Indicates there are two possible solutions, one involving addition and the other involving subtraction.
• The numerator $-b$: Represents the inversion of the linear coefficient.
• The discriminant $\sqrt{b^2-4ac}$:
  • The expression under the square root, $b^2-4ac$, is called the discriminant. It determines the nature of the roots of the quadratic equation (real and distinct, real and repeated, or complex conjugates).

How to Use the Quadratic Formula

1. Write the equation in standard form $a{x}^2+bx+c=0$.
2. Identify the coefficients $a$, $b$, and $c$.
3. Substitute the values of $a$, $b$, and $c$ into the quadratic formula.
4. Calculate the discriminant $b^2-4ac$ to determine the nature of the roots.
5. Solve the formula for $x$, simplifying as necessary.

Example Using the Quadratic Formula

Solve the quadratic equation $2x^2-4x-6=0$ using the quadratic formula.

1. Identify the coefficients: $a=2$, $b=-4$, $c=-6$.
2. Substitute into the formula:

$x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(2)(-6)}}{2(2)}$

$x=\frac{4\pm \sqrt{16+48}}{4}$

$x=\frac{4\pm \sqrt{64}}{4}$

$x=\frac{4\pm 8}{4}$

3. Calculate both possible solutions:

$x=\frac{4+8}{4}=3$ $x=\frac{4-8}{4}=-1$

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.

Examples of Quadratics

Example 1:

Problem: Solve the quadratic equation $x^2-7x+10=0$ .

Solution:

1. Write the equation in standard form: $x^2-7x+10=0$.
2. Find two numbers that multiply to 10 and add to -7.
   • The numbers are -5 and -2 because (-5) * (-2) = 10 and (-5) + (-2) = -7.
3. Rewrite the equation using these numbers: $x^2-5x-2x+10=0$.
4. Factor by grouping.
   • Group terms: $(x^2-5x)+(-2x+10)=0$.
   • Factor out the common terms: $x(x-5)-2(x-5)=0$.
5. Factor out the common binomial: $(x-5)(x-2)=0$.
6. Set each factor equal to zero: $x-5=0$ or $x-2=0$.
7. Solve for $x$:
   • $x=5$
   • $x=2$

Answer: The solutions are $x=5$ and $x=2$.

Example 2:

Problem: Solve the quadratic equation $x^2+6x-7=0$ by completing the square.

Solution:

1. Write the equation in standard form: $x^2+6x-7=0$.
2. Move the constant term to the other side: $x^2+6x=7$.
3. To complete the square, add $(\frac{6}{2}{)}^2=9$ to both sides: $x^2+6x+9=7+9$.
4. The left side is a perfect square trinomial: $(x+3{)}^2=16$.
5. Take the square root of both sides: $x+3=\pm 4$.
6. Solve for $x$:
   • $x+3=4$ gives $x=1$.
   • $x+3=-4$ gives $x=-7$.

Answer: The solutions are $x=1$ and $x=-7$.

Example 3:

Problem: Use the quadratic formula to solve the equation $3x^2-2x-1=0$.

Solution:

1. Identify the coefficients $a=3$, $b=-2$, and $c=-1$ in the standard form $a{x}^2+bx+c=0$.
2. Plug these coefficients into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(3)(-1)}}{2(3)}$

$x=\frac{2\pm \sqrt{4+12}}{6}$

$x=\frac{2\pm \sqrt{16}}{6}$

$x=\frac{2\pm 4}{6}$

3. Solve for $x$ using the two possible signs:
   • For the positive sign: $x=\frac{2+4}{6}=\frac{6}{6}=1$.
   • For the negative sign: $x=\frac{2-4}{6}=\frac{-2}{6}=-\frac{1}{3}$.

Answer: The solutions are $x=1$ and $x=-\frac{1}{3}$.

How to Solve Quadratic Equations?

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 Method

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:

1. Write the quadratic equation in standard form: $a{x}^2+bx+c=0$.
2. Find two numbers that multiply to give $ac$ (the product of the coefficient of $x^2$ and the constant term) and add to give $b$ (the coefficient of $x$).
3. Rewrite the middle term $bx$ using the two numbers found.
4. Factor by grouping, which involves dividing the quadratic into two parts and factoring out the greatest common factor from each.
5. Set each factor equal to zero and solve for $x$.

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 $x$:

$x-2=0\text{or}x-3=0$

$x=2\text{or}x=3$

Completing the Square Method

Completing the square involves creating a perfect square trinomial from the quadratic equation, which can then easily be solved.

Steps for completing the square:

1. Write the quadratic equation in the form $a{x}^2+bx+c=0$, and divide all terms by $a$ (if $a\ne 1$).
2. Move the constant term to the opposite side of the equation.
3. Add and subtract $(b/(2a){)}^2$ to the left side to complete the square.
4. Factor the perfect square trinomial on the left side.
5. Take the square root of both sides, remembering to consider both the positive and negative roots.
6. Solve for $x$ by isolating it on one side of the equation.

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:

$x-2=\pm 1$

$x=2\pm 1$

$x=3\text{or}x=1$

Quadratic Formula Method

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:

1. Write down the quadratic equation in standard form.
2. Identify the coefficients $a$, $b$, and $c$.
3. Substitute these values into the quadratic formula.
4. Simplify and solve for $x$.

Example: Solve $x^2+6x-7=0$ using the quadratic formula. Here, $a=1$, $b=6$, and $c=-7$. Substitute these into the formula:

$x=\frac{-6\pm \sqrt{6^2-4(1)(-7)}}{2(1)}$

$x=\frac{-6\pm \sqrt{36+28}}{2}$

$x=\frac{-6\pm \sqrt{64}}{2}$

$x=\frac{-6\pm 8}{2}$

$x=-7\text{or}x=1$

Graphing Method

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:

1. Write the quadratic equation in standard form.
2. Determine the vertex of the parabola and the axis of symmetry.
3. Plot the vertex and a few additional points on the graph.
4. Draw the parabola through the points.
5. Identify the x-intercepts (if any), which are the solutions to the equation.

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.

Range of Quadratic Equations

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:

$f(x)=a{x}^2+bx+c,$

where $a$, $b$, and $c$ are constants, and $a\ne 0$.

Here's how to determine the range of a quadratic function:

1. Identify the Orientation of the Parabola:

   • If $a>0$, the parabola opens upward.
   • If $a<0$, the parabola opens downward.

2. 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:

   • $h=\frac{-b}{2a}$
   • $k=f(h)=a(h{)}^2+b(h)+c$

3. Determine the Range:

   • If the parabola opens upwards ($a>0$), the range starts at the y-coordinate of the vertex and goes to positive infinity, because as $x$ moves away from the vertex in either direction, the y-values keep increasing. Thus, the range is $[k,\mathrm{\infty })$, where $k$ is the y-value of the vertex.
   • If the parabola opens downwards ($a<0$), the range goes from negative infinity up to the y-coordinate of the vertex, because as $x$ moves away from the vertex, the y-values decrease. Thus, the range is $(-\mathrm{\infty },k]$, where $k$ is the y-value of the vertex.

Example Calculation

Consider the quadratic function:

$f(x)=2x^2-4x+1.$

1. Since $a=2$ is positive, the parabola opens upward.

2. To find the vertex:

   • $h=\frac{-(-4)}{2(2)}=\frac{4}{4}=1$
   • $k=f(1)=2(1{)}^2-4(1)+1=2-4+1=-1$ So the vertex is at $(1,-1)$.

3. With the parabola opening upward and the vertex being $(1,-1)$, the range of the function is $[-1,\mathrm{\infty })$.

Therefore, for this quadratic function, the output values $y$ can be any real number that is $-1$ or greater.

Solved Word Problems on Quadratic Equations

Problem 1: Projectile Motion

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.

Problem 2: Optimal Revenue

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.

Problem 3: Area of a Garden

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.

Applications of Quadratic Equations

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:

Physics: Projectile Motion

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:

$y(t)=v_{0y}t-\frac{1}{2}g{t}^2+h_0,$

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.

Engineering: Structural Design

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).

Economics: Revenue and Cost Optimization

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.

Optics: Design of Lenses and Mirrors

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.

Ballistics: Trajectories

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.

Biology: Population Dynamics

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.

Sports: Predicting Paths

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.

Chemistry: Reaction Rates

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.

Architecture: Parabolic Structures

Quadratic equations describe parabolic arches and other architectural elements that are not only aesthetically pleasing but also structurally efficient, distributing weight and resistance evenly.

Entertainment: Computer Graphics

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.

Acoustics: Sound Reflections

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.

Mathematics: Finding Integers

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.

Practice Questions

1. Solve $x^2+x-12=0$.
2. Solve $2x^2-7x=3$ by completing the square.
3. Find the range of $y=-3x^2+6x-2$.

Frequently Asked Questions and Answer on Quadratics

What exactly is a quadratic equation?

A quadratic equation is a second-degree polynomial equation in a single variable x with the general form:

$a{x}^2+bx+c=0,$

where $a$, $b$, and $c$ are constants with $a\ne 0$.

How do you know if an equation is quadratic?

An equation is quadratic if it can be written in the form $a{x}^2+bx+c=0$ and $a$ is not equal to zero. The highest power of the variable x is 2.

Can a quadratic equation have more than two solutions?

No, a quadratic equation in one variable has at most two distinct solutions. These solutions can be real or complex numbers.

Why can't the 'a' in the standard form of a quadratic equation be zero?

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 $a{x}^2$ term.

What is the discriminant, and what does it tell us about a quadratic equation?

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:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root (a repeated root).
• If the discriminant is negative, there are no real roots; instead, there are two complex conjugate roots.

How is the vertex of a parabola related to the quadratic equation?

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.

What are the real-life applications of quadratic equations?

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).

How is completing the square used to solve quadratic equations?

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.

Why is the quadratic formula important?

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.