/*! 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 82 Solve each equation. $$\sqrt[3... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 82

Solve each equation. $$\sqrt[3]{2-w}=\sqrt[3]{2 w-28}$$

Short Answer

Expert verified
The solution is \( w = 15 \).

Step by step solution

01

- Set the expressions inside the cube roots equal

Since the cube roots are equal, set the expressions inside them equal to each other: \ \(2 - w = 2w - 28\)
02

- Rearrange the equation

Rearrange the equation to get all the terms involving the variable on one side. Move all terms involving \(w\) to one side and constants to the other: \ \(2 - 2w = -28\)
03

- Solve for \(w\)

Solve the resulting linear equation by isolating \(w\): \ \(2 + 28 = 2w\) \ \(30 = 2w\) Divide both sides by 2: \ \(w = 15\)
04

- Verify the solution

Substitute \(w = 15\) back into the original equation to ensure both sides are equal: \ \(\sqrt[3]{2 - 15} = \sqrt[3]{2(15) - 28}\) \ \(\sqrt[3]{-13} = \sqrt[3]{2}\) Both sides are equal, confirming that \(w = 15\) is correct.

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.

Cube Root Equations
A cube root equation involves the cube root of a variable or expression. The cube root of a number is a value that, when multiplied by itself three times, gives the original number.
For example, the cube root of 8 is 2, because 2\textsuperscript{3} = 8.
In the given problem, the cube root equation is \( \sqrt[3]{2 - w} = \sqrt[3]{2w - 28} \). To solve this, we use the property: if \sqrt[3]{A} = \sqrt[3]{B}, then A = B.
This allows us to set the expressions inside the cube roots equal to each other.
  • Step 1: Set the expressions equal: \(2 - w = 2w - 28\)
  • Step 2: Rearrange to isolate the variable.
Linear Equations
A linear equation is an equation where the highest power of the variable is 1. It has the form \(Ax + B = 0\), where A and B are constants, and x is the variable.
In our problem, once we set the expressions inside the cube roots equal to each other, we get a linear equation: \(2 - w = 2w - 28\).
We solve it by isolating the variable (w):
  • Step 3: Move variable terms to one side and constant terms to the other:
    • \(2 - 2w = -28\)
  • Add 28 to both sides to maintain equality:
    • \(2 + 28 = 2w\)
  • Combine like terms:
    • \(30 = 2w\)
  • Divide both sides by 2:
    • \(w = 15\)

This process transforms the equation step by step until the solution is found.
Equation Verification
Equation verification is the process of confirming that a solution satisfies the original equation.
After solving for w, we substitute it back into the original equation to ensure both sides are equal.
Substitute w = 15 back into the cube root equation:
  • \(\sqrt[3]{2 - 15} = \sqrt[3]{2(15) - 28}\)
  • Simplify both expressions:
    • \(\sqrt[3]{-13} = \sqrt[3]{2}\)

Both sides of the equation are indeed equal, confirming that our solution w = 15 is correct.
This is a crucial step in solving equations to ensure no errors were made during the process. Following these steps verifies the solution and confirms its correctness.

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 equation. $$x^{-3}=8$$

Solve each equation. $$2 x^{2}+3=7$$

Simplify. $$\frac{4-\sqrt{28}}{6}$$

Simplify. $$\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{\sqrt{5}}{3 \sqrt{2}}$$

Solve each equation. $$t^{-1 / 2}=9$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Applied Mathematics

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.