Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 191

Solution set of equation \(\mathrm{x}=\sqrt{[12+\sqrt{\\{} 12+\sqrt{(} 12)\\}] \ldots \ldots . . \text { up to }}\) \(\infty\) is (a) 4 (b) \(-4\) (c) 3 (d) \(-3\)

Short Answer

Expert verified
The solution set of the infinite nested radical equation \(x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\) is (a) 4.

Step by step solution

01

Set up the equation

First, write the given infinite nested equation: \[x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\] Since this is an infinite nested radical, we can assume that the next term will have the same structure as the whole expression: \[x = \sqrt{12+x}\] #Step 2: Solve for x#
02

Solve for x

To solve for x, square both sides of the equation: \[x^2 = 12+x\] Rearrange the equation to have a quadratic equation: \[x^2 - x - 12 = 0\] Now, we can factor the left side of the equation: \[(x-4)(x+3) = 0\] This gives us two possible solutions for x: \[x = 4, \; -3\] #Step 3: Verify the solutions#
03

Verify the solutions

We must now check if these solutions are valid in the context of our original infinite nested radical equation. If we substitute x = 4 back into the equation, we get: \[4 = \sqrt{12+4}\] \[4 = \sqrt{16}\] \[4 = 4\] This solution is valid. Now, let's substitute x = -3: \[-3 = \sqrt{12-3}\] \[-3 = \sqrt{9}\] However, \(\sqrt{9}\) = 3, not -3, therefore, x = -3 is not a valid solution. After verifying the solutions, we find that the only valid solution from the answer choices is: #Answer# (a) 4

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If the sum of the two roots of the equation \([1 /(x+a)]+[1 /(x+b)]=(1 / k)\) is zero then their Product is (a) \(\overline{(1 / 2)\left(a^{2}+b^{2}\right)}\) (b) \(\\{(-1) / 2\\}(a+b)^{2}\) (c) \([\\{a+b\\} / 2]^{2}\) (d) None

If \(\alpha, \beta\) are roots of \(8 \mathrm{x}^{2}-3 \mathrm{x}+27=0\) then \(\left(\alpha^{2} / \beta\right)^{(1 / 3)}+\left(\beta^{2} / \alpha\right)^{(1 / 3)}=\) (a) \((1 / 3)\) (b) \((7 / 2)\) (c) 4 (d) \((1 / 4)\)

The solution set of equation \([(\mathrm{x}-\mathrm{a}) /(\mathrm{x}-\mathrm{b})]+\) \([(x-b) /(x-a)]=(a / b)+(b / a),(a \neq b)\) is (a) \(\\{\mathrm{a}-\mathrm{b}, 0\\}\) (b) \(\\{(\mathrm{a} / \mathrm{b}), 0\\}\) (c) \(\\{a+b, 0\\}\) (d) \(\\{\mathrm{ab}, 0\\}\)

If the roots of the equation \(x^{2}-b x+c=0\) be two consecutive integers then \(b^{2}-4 c=\) (a) \(-2\) (b) \(-3\) (c) 3 (d) 1

In \(\triangle \mathrm{ABC}, \mathrm{m} \angle \mathrm{C}=(\pi / 2)\) If \(\tan (\mathrm{A} / 2)\) and \(\tan (\mathrm{B} / 2)\) are the roots of \(a x^{2}+b x+c=0, a \neq 0\) then (a) \(b=a+c\) (b) \(b=c\) (c) \(c=a+b\) (d) \(a=b+c\)
See all solutions

Recommended explanations on English Textbooks

Cues and Conventions

Read Explanation

Essay Plan

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Phonology

Read Explanation

Free Response Essay

Read Explanation

English Grammar Summary

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.

Sign-up for free