# 2.3: Quadratic Equations with Applications

- Page ID
- 152030

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)After completing this section, you should be able to:

- Multiply polynomials.
- Factor trinomials.
- Solve quadratic equations by factoring.
- Solve quadratic equations using the quadratic formula.
- Solve real-world applications modeled by quadratic equations.

In this section, we will discuss quadratic equations. There are several real-world scenarios that can be represented by the graph of a quadratic equation. Think of the Gateway Arch in St. Louis, Missouri. Both ends of the arch are 630 feet apart and the arch is 630 feet tall. You can plot these points on a coordinate system and create a parabola to graph the quadratic equation.

Before we begin working with solving quadratic equations, let's do some quick review. A **monomial **is a term of the form $a{x}^{m}$ where $a$ is any real number and $m$ is a whole number. Some examples of monomials are 3x, $5{x}^{7}$, and 6. We can add (or subtract) several monomials together to form a **polynomial**. Some examples of polynomials are $3x-2$, $-123$, ${y}^{2}+4y-\frac{1}{3}$, and $2{x}^{7}+5x$. Polynomials can be named more specifically using the number of terms they have. If a polynomial has two terms, it is called a **binomial** (examples: $3x-2$ or $2{x}^{7}+5x$). If a polynomial has three terms, it is called a **trinomial** (example:${y}^{2}+4y-\frac{1}{3}$).

### Multiply Polynomials

Distribution is the key to multiplying polynomials. You are used to distributing a number when solving equations. For example 2(x+3) can be rewritten as 2x + 6. Let's look at what happens if we add a variable to the monomial outside the parentheses.

Multiply 3x(2x + 1).

**Answer**

3x (2x + 1) | |

Distribute the 3x. | $3x\cdot 2x+3x\cdot 1$ |

Simplify. | $6{x}^{2}+3x$ |

Multiply 4x (5x - 7).

Now let's try multiplying two binomials. You may have learned the Box Method or FOIL in a previous course to help you remember the process for multiplying two binomials. Both of these are really just distributing each term from the first binomial to each term of the second binomial.

Multiply (2x - 5)(x + 6)

###### Solution

(2x - 5)(x + 6) | |

Distribute the 2x to each term of the second binomial. Distribute the -5 to each term of the second binomial. |
$2x\cdot x+2x\cdot 6-5\cdot x-5\cdot 6$ |

Multiply the terms. | $2{x}^{2}+12x-5x-30$ |

Simplify. | $2{x}^{2}+7x-30$ |

Multiply (x - 8)(4x + 3).

You may remember factoring trinomials of the form ${x}^{2}+bx+c$ in previous courses. You should look for two numbers that add to ${b}^{}$ and multiply to ${c}^{}$. Let's look at an example as a review.

Factor ${x}^{2}-16x+63$

**Answer**

Here we are looking for two numbers that add to -16 and multiply to 63.

Let's start by thinking through the factors of 63. We have 1 and 63, 3 and 21, or 7 and 9.

Which pair of factors adds to -16? 7 and 9 add to 16, but we are looking for -16.

If we make both the 7 and 9 negative, we have -7 + -9 = -16 and -7 ⋅ -9=63.

So the trinomial factors into (x-7)(x-9).

Factor ${x}^{2}-5x-14$

Now that we've spent some time reviewing, let's look at quadratic equations.

A quadratic equation has a polynomial where the highest exponent is 2, and is often written with the polynomial equal to zero. It must be able to be written in the form $a{x}^{2}+bx+c=0$, where ${a}^{}$, ${b}^{}$, and ${c}^{}$ are real numbers.

Since we are working with equations now, we will want to solve them. This means we are trying to find the value(s) of x that make the equation true.

### Solving Quadratic Equations by Factoring

One way to solve a quadratic equation is by factoring. We will use the Zero Product Property that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

#### Solving a Quadratic Equation by Factoring

Solve ${x}^{2}+2x-15=0$.

**Answer**

Factor the trinomial on the left hand side. What two numbers add to 2 and multiply to -15? |
Factors of -15 are -1 and 15, 1 and -15, -3 and 5, or 3 and -5. The pair that adds to 2 is -3 and 5. So the trinomial factors into (x - 3)(x + 5). |

Rewrite the equation with the factored form. | (x - 3)(x + 5) = 0 |

Apply the Zero Product Property. | x - 3 = 0 or x + 5 = 0 |

Solve each equation. | x = 3 or x = -5 |

Solve ${x}^{2}+11x+30=0$

### Solving Quadratic Equations Using the Quadratic Formula

The other method we will look at for solving quadratic equations is the quadratic formula. This method works for all quadratic equations, even quadratic equations we could not factor!

The solutions to an equation of the form $a{x}^{2}+bx+c=0$ can be found using the formula $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Solve using the quadratic formula: $3{x}^{2}-5x+2=0$.

**Answer**

Write the equation in standard form. | Already done! |

Identify the values of a, b, and c. | a = 3, b = -5, and c = 2 |

Plug these values into the quadratic formula. | $x=\frac{5\pm \sqrt{{\mathrm{(-5)}}^{2}-4\cdot 3\cdot 2}}{2\cdot 3}$ |

Simplify. | $x=\frac{5\pm \sqrt{1}}{6}=\frac{5\pm 1}{6}$ |

Solve for each value. | $x=\frac{5+1}{6}=1$ or $x=\frac{5-1}{6}=\frac{4}{6}=\frac{2}{3}$ |

Solve using the quadratic formula: $4{x}^{2}+8x+7=4$.

### Solving Real-World Applications Modeled by Quadratic Equations

There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems:

**Step 1:** Read the problem. Make sure all the words and ideas are understood.

**Step 2:** Identify what we are looking for.

**Step 3:** Name what we are looking for. Choose a variable to represent that quantity.

**Step 4:** Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.

**Step 5:** Solve the equation using good algebra techniques.

**Step 6:** Check the answer in the problem and make sure it makes sense.

**Step 7:** Answer the question with a complete sentence.

The product of two consecutive integers is 132. Find the integers.

**Answer**-
**Step 1:**Read the problem.**Step 2:**Identify what we are looking for.We are looking for two consecutive integers.

**Step 3:**Name what we are looking for. Choose a variable to represent that quantity.We are looking for two integers that are consecutive. If we call the first one $x$, then the next consecutive integer would be $x+1$.

**Step 4:**Translate into an equation.The product of the consecutive integers is 132. So $x(x+1)=132$.

**Step 5:**Solve the equation using good algebra techniques.Write the equation in standard form. ${x}^{2}+x=132$

${x}^{2}+x-132=0$

Solve the equation by factoring.

(You could also use the quadratic formula here.)

(x + 12)(x - 11) = 0

x + 12 = 0 or x - 11 = 0

x = -12 or x = 11

**Step 6:**Check the answer in the problem and make sure it makes sense.If x = -12, then the next consecutive integer would be -11. -12 ⋅ -11 = 132, so that solution works!

If x = 11, then the next consecutive integer would be 12. 11 ⋅ 12 = 132, so that solution also works!

**Step 7:**Answer the question with a complete sentence.-12 and -11 are consecutive integers whose product is 132. 11 and 12 are another set of consecutive integers whose product is 132.

Two consecutive integers have a product of 72. What are the integers?

Quadratic equations can be used to model projectile motion. When you launch an object into the air, it will go up for awhile, but eventually gravity will take over, and the object will start falling. The path of the object is a parabola, similar to the St. Louis arch we looked at during the start of this section.

A baseball player hits a ball into the air. The height (in feet) t seconds after the ball is hit can be given by the equation $h=\mathrm{-}16{t}^{2}+47t+3$. Assuming no one catches the ball, how long will it be in the air?

###### Solution

If we want to know how long the ball is in the air, we want to know at what time the ball will hit the ground. When the ball is on the ground, the height is 0 feet. So let's set the height equal to zero and solve the equation.

$0=\mathrm{-}16{t}^{2}+47t+3$.

We need to solve this equation. We could try factoring it, or we could use the quadratic equation. Let's try the quadratic equation this time.

$t=\frac{-47\pm \sqrt{{47}^{2}-4\cdot -16\cdot 3}}{2\cdot -16}$

$t=\frac{-47\pm \sqrt{2401}}{-32}=\frac{-47\pm 49}{-32}$

So $t=\frac{-47+49}{-32}=\frac{2}{-32}=-0.0625$ or $t=\frac{-47-49}{-32}=\frac{-96}{-32}=3$

So, we get that the ball is on the ground at -0.0625 seconds and at 3 seconds. The -0.0625 does not make sense in the problem, since time cannot be negative. So, we will say that the ball lands on the ground after 3 seconds.

A child throws a rock into a river. The height of the rock t seconds after being thrown can be given using $h=\mathrm{-}16{t}^{2}-12t+10$. How long will it take before the rock lands in the river?