/*! 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 152 Explain how to solve \(x^{2}+6 x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 152

Explain how to solve \(x^{2}+6 x+8=0\) by completing the square.

Short Answer

Expert verified
The solutions to the equation \(x^{2}+6x+8=0\) by the method of completing the square are \(x=0\) and \(x=-6\).

Step by step solution

01

Identify a, b, and c from the quadratic equation

In a general quadratic equation \(ax^{2}+bx+c=0\), the value of a is the coefficient of \(x^{2}\), b is the coefficient of x, and c is the constant. Here, for the equation \(x^{2}+6x+8=0\), the values are \(a=1, b=6, c=8\).
02

Rewrite the equation in the form \((x-m)^{2}=n\)

To complete the square, the given equation needs to be in the form of \((x-m)^{2}=n\), where m is half the coefficient of x and n is a constant. Rewrite the equation in that format. The equation becomes \((x+3)^{2}-9=0\), where 3 is half of 6, \(3^{2} = 9\) is subtracted to balance the added square in the left hand side of the equation.
03

Simplify and solve for x

Solving the equation \((x+3)^{2}-9=0\), we get \((x+3)^{2}=9\). Taking the square root of both sides of the equation, we obtain \(x+3= 卤\sqrt{9}\). Hence, \(x=-3 卤\sqrt{9}\), which gives the solutions \(x=-3+3=-0\) and \(x=-3-3=-6\).

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!

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

The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

The rectangular swimming pool in the figure shown measures 40 feet by 60 feet and is surrounded by a path of uniform width around the four edges. The perimeter of the rectangle formed by the pool and the surrounding path is 248 feet. Determine the width of the path.

The bar graph shows median yearly earnings of full-time workers in the United States for people 25 years and over with a college education, by final degree earned. Exercises \(3-4\) are based on the data displayed by the graph. (graph can't copy) The median yearly salary of an American whose final degree is a master's is \(\$ 70\) thousand less than twice that of an American whose final degree is a bachelor's. Combined, two people with each of these educational attainments earn \(\$ 173\) thousand. Find the median yearly salary of Americans with each of these final degrees.

You need to rent a rug cleaner. Company A will rent the machine you need for \(\$ 22\) plus \(\$ 6\) per hour. Company \(B\) will rent the same machine for \(\$ 28\) plus \(\$ 4\) per hour. After how many hours of use will the total amount spent at each company be the same? What will be the total amount spent at each company?

A 20 -foot ladder is 15 feet from a house. How far up the house, to the nearest tenth of a foot, does the ladder reach?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Calculus

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.