/*! 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 85 Solve each equation. $$6(x+2)^... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 85

Solve each equation. $$6(x+2)^{4}-11(x+2)^{2}=-4$$

Short Answer

Expert verified
The solutions are \( x = -2 \pm \sqrt{\frac{4}{3}} \) and \( x = -2 \pm \sqrt{\frac{1}{2}} \).

Step by step solution

01

Introduce a substitution

Let \( y = (x + 2)^{2} \). This substitution will simplify the equation. Now, rewrite the given equation using \( y \).
02

Rewrite the equation

Substituting \( y \) into the equation, we get: \( 6y^{2} - 11y = -4 \).
03

Move all terms to one side

Add 4 to both sides to set the equation to zero: \( 6y^{2} - 11y + 4 = 0 \).
04

Factor the quadratic equation

Factorize \( 6y^{2} - 11y + 4 \) into two binomials: \( (3y-4)(2y-1) = 0 \).
05

Solve for \( y \)

Set each factor equal to zero and solve for \( y \):
06

Solve \( 3y-4 = 0 \)

Solving \( 3y-4 = 0 \) yields \( y = \frac{4}{3} \).
07

Solve \( 2y-1 = 0 \)

Solving \( 2y-1 = 0 \) yields \( y = \frac{1}{2} \).
08

Substitute back \( y = (x+2)^{2} \)

Now, substitute back \( y \) into the original substitution. So, \( (x+2)^{2} = \frac{4}{3} \) or \( (x+2)^{2} = \frac{1}{2} \).
09

Solve for \( x \) from \( (x+2)^{2} = \frac{4}{3} \)

Taking the square root of both sides, we get \( x + 2 = \pm \sqrt{\frac{4}{3}} \). This simplifies to two solutions: \( x = -2 \pm \sqrt{\frac{4}{3}} \).
10

Solve for \( x \) from \( (x+2)^{2} = \frac{1}{2} \)

Taking the square root of both sides, we get \( x + 2 = \pm \sqrt{\frac{1}{2}} \). This simplifies to two solutions: \( x = -2 \pm \sqrt{\frac{1}{2}} \).
11

Write the final solutions

The final solutions for the given equation are: \( x = -2 + \sqrt{\frac{4}{3}} \), \( x = -2 - \sqrt{\frac{4}{3}} \), \( x = -2 + \sqrt{\frac{1}{2}} \), and \( x = -2 - \sqrt{\frac{1}{2}} \).

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.

Substitution Method
The substitution method is a powerful technique to simplify complicated equations. By introducing a new variable, we can turn a complex problem into a simpler one. For instance, in the problem, we let \( y = (x + 2)^{2} \). This changes the original equation into a quadratic form: \( 6y^{2} - 11y = -4 \). The new equation is easier to handle because it is a standard quadratic. This method is especially useful when dealing with polynomial or rational equations. By replacing parts of the equation with a new variable, we simplify the original expression, making it easier to solve.
Factoring Quadratics
Factoring quadratics is a technique used to break down a quadratic equation into simpler, solvable parts. After substitution, the equation became \( 6y^{2} - 11y + 4 = 0 \). We then factored it into \( (3y-4)(2y-1) = 0 \). Factoring makes it easy to find the solutions: set each factor to zero and solve for \( y \). This process involves:
  • Finding two numbers that multiply to the constant term (in this case, 4) and add to the coefficient of the middle term (in this case, -11).
  • Using these numbers to break the middle term and factor by grouping.
Once the quadratic is factored into binomials, solving them individually gives the possible values for \( y \).
Completing the Square
Completing the square is a method to solve quadratic equations by transforming them into a perfect square trinomial. Although this method was not directly used in our problem, it's another way to solve quadratics. To complete the square:
  • Ensure the quadratic equation is in the form \( ax^2 + bx + c = 0 \).
  • Move the constant term to the other side: \( ax^2 + bx = -c \).
  • If \( a \) is not 1, divide the entire equation by \( a \).
  • Add and subtract \( \left( \frac{b}{2a} \right)^2 \) inside the equation.
  • Rewrite the left side as a square of a binomial.
This method transforms the quadratic into an equation of the form \( (x+p)^2 = q \), which can be easily solved by taking the square root.
Radical Expressions
Radical expressions involve numbers and variables under the radical sign. In our solved problem, after substituting back, we ended up with \( (x + 2)^{2} = \frac{4}{3} \) and \( (x + 2)^{2} = \frac{1}{2} \). Solving these required taking the square root of both sides:
  • \( x + 2 = \pm \sqrt{\frac{4}{3}} \)
  • \( x + 2 = \pm \sqrt{\frac{1}{2}} \)
This step introduced radical expressions. Taking the square root of a fraction involves taking the square root of both the numerator and the denominator individually. Simplifying these square roots gave us the final solutions. Understanding how to work with radicals is crucial as they often appear in algebra and calculus.

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

Find each product. Write the answer in standard form. $$i(3-4 i)(3+4 i)$$

Solve each rational inequality. Write each solution set in interval notation.4 $$\frac{(5-3 x)^{2}}{(2 x-5)^{3}}>0$$

Solve each equation or inequality. $$\left|2 x+\frac{1}{3}\right|+1<4$$

Complex numbers are used to describe current I, voltage \(E,\) and impedance \(Z\) (the opposition to current). These three quantities are related by the equation \(E=I Z, \quad\) which is known as Ohm's Law. Thus, if any two of these quantities are known, the third can be found. In each exercise, solve the equation \(E=I Z\) for the remaining value. $$I=10+4 i, E=88+128 i$$

Classroom Ventilation According to the American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (ASHRAE), a nonsmoking classroom should have a ventilation rate of \(15 \mathrm{ft}^{3}\) per min for each person in the room. (a) Write an equation that models the total ventilation \(V\) (in cubic feet per hour) necessary for a classroom with \(x\) students. (b) A common unit of ventilation is air change per hour (ach). 1 ach is equivalent to exchanging all of the air in a room every hour. If \(x\) students are in a classroom having volume \(15,000 \mathrm{ft}^{3},\) determine how many air exchanges per hour \((A)\) are necessary to keep the room properly ventilated. (c) Find the necessary number of ach ( \(A\) ) if the classroom has 40 students in it. (d) In areas like bars and lounges that allow smoking, the ventilation rate should be increased to \(50 \mathrm{ft}^{3}\) per min per person. Compared to classrooms, ventilation should be increased by what factor in heavy smoking areas?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.