/*! 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 62 In solving \(\sqrt{2 x-1}+2=x,\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 62

In solving \(\sqrt{2 x-1}+2=x,\) why is it a good idea to isolate the radical term? What if we don't do this and simply square each side? Describe what happens.

Short Answer

Expert verified
If the square root term is not isolated before squaring, the resulting equation becomes far more complicated to solve. Therefore, always isolate the square root before squaring. In the given question, the correct answer turns out to be \( x=1 \).

Step by step solution

01

Isolating square root and squaring

Firstly, isolate the square root term on one side before squaring. The given equation is \( \sqrt{2x - 1} + 2 = x \). Isolate the square root term to look like this: \( \sqrt{2x - 1} = x - 2 \). Now, square both sides. This yields: \(2x - 1 = (x - 2)^2\).
02

Expansion and simplification

Expand the square term on the right side and simplify by placing all terms on the left side. This results in \(2x - 1 = x^2 - 4x + 4\), which simplifies to \(x^2 - 6x + 5 = 0\).
03

Solving the quadratic equation

Solve the resulting quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, the equation can be factored easily into this form: \( (x-5)(x-1) = 0 \). Therefore, the solutions are \( x=1 \) and \( x=5 \).
04

Check solution in original equation

Check these solutions by substituting them in the original equation. If a number makes the original equation true, then it is a valid solution. However, \( x=5 \) simply does not satisfy the original equation. Thus, the only solution to the original equation is \( x=1 \).
05

Procedure without isolating the square root

What if the square root term weren't isolated before squaring? If both sides of the initial equation were squared without isolating the square root, the result would be: \(2x - 1 + 4\sqrt{2x - 1} + 4 = x^2\). This leads to a much more complicated equation involving a square root term that is not isolated, making the equation harder to solve.

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

evaluate each expression, or state that the expression is not a real number. $$\sqrt{0.81}$$

Exercises \(145-147\) show a number of simplifications, not all of which are correct. Enter the left side of each equation as \(y_{1}\) and the right side as \(y_{2} .\) Then use your graphing utility's \(|\) TABLE feature to determine if the simplification is correct. If it is not, correct the right side and use the \([\text { TABLE }]\) feature to verify your simplification. $$\left(x^{-\frac{1}{2}} \cdot x^{\frac{3}{4}}\right)^{-2}=x^{\frac{1}{2}}$$

What is an extraneous solution to a radical equation?

Multiply and simplify. Assume that all variables in a radicand represent positive real numbers and no radicands involve negative quantities raised to even powers. $$\sqrt{2 x^{7}} \cdot \sqrt{12 x^{4}}$$

Divide using synthetic division: $$\left(4 x^{4}-3 x^{3}+2 x^{2}-x-1\right) \div(x+3)$$ (Section \(5.6,\) Example 5 )
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied 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.