/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Solve each equation by factoring... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 37

Solve each equation by factoring. \(x^{2}-16=0\)

Short Answer

Expert verified
The solutions are \( x = 4 \) and \( x = -4 \).

Step by step solution

01

Recognize the Form

Notice that the equation \( x^2 - 16 = 0 \) is a difference of squares. A difference of squares has the form \( a^2 - b^2 = (a - b)(a + b) \).
02

Identify the Perfect Squares

Identify the perfect squares in the expression. Here, \( x^2 \) is the square of \( x \) and \( 16 \) is the square of \( 4 \). Therefore, the equation can be expressed as \((x)^2 - (4)^2 = 0 \).
03

Apply the Difference of Squares Formula

Using the difference of squares formula, factor \( x^2 - 16 \) as \((x - 4)(x + 4) = 0 \).
04

Solve the Factored Equation

Set each factor equal to zero to find the solutions: \( x - 4 = 0 \) or \( x + 4 = 0 \). Solving these gives \( x = 4 \) and \( x = -4 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The concept of the difference of squares is a fundamental algebraic identity. It allows you to factor expressions of the form \(a^2 - b^2\). This specific identity is given as \[(a - b)(a + b)\]It's important because it provides a straightforward way to simplify and solve certain types of equations. You apply this when you recognize both terms as perfect squares.
  • For example, in the equation \(x^2 - 16 = 0\), \(x^2\) is the square of \(x\), and \(16\) is the square of \(4\).
  • The expression transforms as \((x)^2 - (4)^2\).
  • Using the difference of squares formula, it factors into \((x - 4)(x + 4)\).
By understanding this identity, you can quickly factor expressions and find solutions to simple quadratic equations.
Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). In simple terms, they are polynomials of degree two. Solving quadratic equations is a significant part of algebra because these equations frequently appear in various mathematical and real-world applications.
  • The standard form is helpful when you need to quickly identify how to factor or apply special methods to solve it.
  • In the example \(x^2 - 16 = 0\), our equation is already in a simplified quadratic form with \(a = 1\), \(b = 0\), and \(c = -16\).
Sometimes, quadratic equations are complex and require formulas like the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), but when special forms like difference of squares are identified, they provide shortcuts to solutions.
Solving Equations
Solving equations involves finding the values of the variables that make the equation true. With quadratic equations, once you factor an expression, solving becomes a process of setting each factor equal to zero.
  • In \((x - 4)(x + 4) = 0\), you solve by setting each part to zero:
  • \(x - 4 = 0\) gives \(x = 4\).
  • \(x + 4 = 0\) gives \(x = -4\).
This process relies on the zero-product property, which states that if a product equals zero, at least one of the factors must be zero. Therefore, solving quadratic equations by factoring is efficient when expressions can be factored outright or recognized as a difference of squares.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A set of 400 test scores is normally distributed with a mean of 75 and a standard deviation of 8 . What percent of the test scores lie between 67 and 83\(?\)

BINOMIAL DISTRIBUTION For Exercises 34 and \(35,\) use the following information. You can use a TI-83 \(/ 84\) Plus graphing calculator to investigate the graph of a binomial distribution. \(Step 1\) Enter the number of trials in LI. Start with 10 trials. \(Step 2\) Calculate the probability of success for each trial in \(\mathrm{L} 2\) \(Step 3\) Graph the histogram. Use the arrow and ENTER keys to choose \(\mathrm{ON}\) , the histogram, Li as the Xlist, and \(\mathrm{L} 2\) as the frequency. Use the window \([0,10]\) scl: 1 by \([0,0.5]\) scl: 0.1 What type of distribution does the binomial distribution start to resemble as \(n\) increases?

For Exercises \(12-21,\) find the margin of sampling error to the nearest percent. $$ p=33 \%, n=1000 $$

REASONING Explain what happens to the margin of sampling error when the size of the sample \(n\) increases. Why does this happen?

MEDIA For Exercises 6 and 7 , use the following information. A survey found that 57\(\%\) of consumers said they will not have any debt from holiday spending. Suppose the survey had a margin of error of 3\(\%\) . What does the 3\(\%\) indicate about the results?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.