/*! 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 11 Use a check to determine whether... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 11

Use a check to determine whether \(-2+\sqrt{2}\) is a solution of \(x^{2}+4 x+2=0\)

Short Answer

Expert verified
Yes, \(-2 + \sqrt{2}\) is a solution.

Step by step solution

01

Understand the Equation

The given equation is a quadratic equation, \(x^2 + 4x + 2 = 0\). We need to determine if \(-2 + \sqrt{2}\) satisfies this equation by substituting it into the equation and checking if it simplifies to zero.
02

Substitute the Value

Replace \(x\) in the equation with \(-2 + \sqrt{2}\): \((-2 + \sqrt{2})^2 + 4(-2 + \sqrt{2}) + 2 = 0\). We will evaluate this expression step by step.
03

Simplify the Squared Term

Calculate \((-2 + \sqrt{2})^2\). This is \((-2)^2 + 2(-2)(\sqrt{2}) + (\sqrt{2})^2 = 4 - 4\sqrt{2} + 2\). The result is \(6 - 4\sqrt{2}\).
04

Evaluate the Linear Term

Calculate \(4(-2 + \sqrt{2})\). This is \(4 \times -2 + 4 \times \sqrt{2} = -8 + 4\sqrt{2}\).
05

Combine and Simplify All Terms

Add the results from step 3, step 4, and the constant in the polynomial: \((6 - 4\sqrt{2}) + (-8 + 4\sqrt{2}) + 2\). Simplifying, we have: \(6 - 8 + 2 + (-4\sqrt{2} + 4\sqrt{2}) = 0\).
06

Confirm the Solution

The entire expression simplifies to 0, which confirms that \(-2 + \sqrt{2}\) is indeed a solution to the equation \(x^2 + 4x + 2 = 0\).

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.

Checking Solutions
To check if a particular value is a solution to a quadratic equation, such as \(x^2 + 4x + 2 = 0\), we substitute the value into the equation and see if the equation holds true, i.e., simplifies to zero.
In this problem, the quantity we're checking is \(-2 + \sqrt{2}\). To verify if it's a solution, we replace every instance of \(x\) in the equation with this value.
  • If substituting in results in an equation that equals zero, \(-2 + \sqrt{2}\) is a solution.
  • If it does not equal zero, then \(-2 + \sqrt{2}\) is not a solution.
Checking solutions is a crucial skill in algebra because it confirms whether the proposed solution actually resolves the equation, ensuring accuracy in solving algebraic problems like quadratic equations.
Substitution Method
The substitution method involves replacing a variable with a particular value to test its validity as a solution. In our problem, we replaced \(x\) with \(-2 + \sqrt{2}\) in the quadratic equation.
By substituting \(x = -2 + \sqrt{2}\) into the equation \(x^2 + 4x + 2 = 0\), the equation becomes \((-2 + \sqrt{2})^2 + 4(-2 + \sqrt{2}) + 2\). This requires calculating each term separately and then combining the results.
  • This way, you can test a suspected solution directly in the equation.
  • It provides a systematic way to test specific values.
  • Substitution is often used when one suspects that a particular value is a solution but needs confirmation.
The substitution method is fundamental in algebra as it allows the application of hypothetical values to see if they solve a given equation or not.
Polynomial Simplification
Simplifying polynomial expressions is an important step in validating solutions in algebra, especially when checking if a value resolves an equation properly.
In the example provided, polynomial simplification involved breaking down the expression \((-2 + \sqrt{2})^2 + 4(-2 + \sqrt{2}) + 2\) into simpler parts:
  • Calculate \((-2 + \sqrt{2})^2\) which resulted in \(6 - 4\sqrt{2}\).
  • Next, compute \(4(-2 + \sqrt{2})\), giving \(-8 + 4\sqrt{2}\).
  • Finally, combine the terms: \(6 - 8 + 2 + (-4\sqrt{2} + 4\sqrt{2})\), which simplifies to \(0\).
The simplification demonstrates that the sum resolver is \(0\), confirming that \(-2 + \sqrt{2}\) is a solution. Polynominal simplification helps clear up complex expressions and checks if an algebraic solution holds true when substituted into the equation.

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

Solve each inequality. Write the solution set in interval notation and graph it. $$ x^{2}+6 x \geq-9 $$

Solve each equation. Approximate the solutions to the nearest hundredth when appropriate. $$ m^{2}-7 m+3=0 $$

After peaking in \(1970,\) school enrollment in the United States fell during the 1970 s and 1980 s. The total annual enrollment (in millions) in U.S. elementary and secondary schools for the years \(1975-1996\) is given by the model \(E(x)=0.058 x^{2}-1.162 x+50.604\) where \(x=0\) corresponds to \(1975, x=1\) corresponds to \(1976,\) and so on. For this period, when was enrollment the lowest? What was the enrollment?

Solve each inequality. Write the solution set in interval notation and graph it. $$ x^{2}+4 x+4>0 $$

Simplify each expression. Assume all variables represent positive numbers. $$ \frac{\sqrt{3}}{\sqrt{50}} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.