/*! 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 40 Find all the real-number roots o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all the real-number roots of each equation. In each case, give an exact expression for the root and also (where appropriate) a calculator approximation rounded to three decimal places. $$\log _{2}\left(2 x^{2}+4\right)=5$$

Short Answer

Expert verified
The roots are \( x = \pm\sqrt{14} \), approximated as \( x \approx \pm 3.742 \).

Step by step solution

01

Apply the Definition of Logarithms

We start by rewriting the logarithmic equation in its exponential form. The equation given is \( \log_{2}(2x^2 + 4) = 5 \). In exponential form, this is: \[ 2x^2 + 4 = 2^5 \] Calculate \( 2^5 \), which equals 32. So, \( 2x^2 + 4 = 32 \).
02

Isolate the Quadratic Term

Next, solve for \( 2x^2 \) by subtracting 4 from both sides of the equation: \[ 2x^2 = 32 - 4 \] This simplifies to: \[ 2x^2 = 28 \].
03

Solve for \( x^2 \)

Divide both sides of the equation by 2 to solve for \( x^2 \): \[ x^2 = \frac{28}{2} \] This results in: \[ x^2 = 14 \].
04

Find the Roots

Now, solve for \( x \) by taking the square root of both sides: \[ x = \pm\sqrt{14} \]. Thus, the exact roots are \( x = \sqrt{14} \) and \( x = -\sqrt{14} \).
05

Calculate Approximations

Use a calculator to determine the decimal approximations of \( \sqrt{14} \) and \( -\sqrt{14} \). \( \sqrt{14} \approx 3.742 \) and \( -\sqrt{14} \approx -3.742 \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real-Number Roots
Real-number roots are the solutions to equations where the variable is not imaginary. For example, in the expression \( x^2 = 14 \), the solutions are real numbers, as they do not involve the imaginary unit \( i \). In simpler terms, real-number roots are the points where a graph intersects the x-axis. They can either be positive, negative, or zero.
When solving any equation, especially ones involving square roots, understanding whether the solutions are real or imaginary is crucial. In this exercise, when we solve \( x^2 = 14 \), we take the square root of both sides to find that the real-number roots are \( x = \sqrt{14} \) and \( x = -\sqrt{14} \). These roots are real because they do not contain any imaginary components.

To summarize, real-number roots indicate where an equation's graph touches or intersects the x-axis, providing tangible solutions that can be verified visually or numerically.
Quadratic Equations
Quadratic equations are polynomial equations of degree 2 and can typically be expressed in the form \( ax^2 + bx + c = 0 \). In our original problem, after applying the definition of logarithms and manipulating the equation, we derived a quadratic equation \( 2x^2 = 28 \).
Quadratic equations can often be solved using several methods, such as factoring, using the quadratic formula, or taking square roots when set up appropriately. In this exercise, we determined \( x^2 = 14 \) and found the solutions by directly taking square roots. This is particularly efficient here because the equation simplifies nicely, but remember that not all quadratic equations are so straightforward.

In general, studying quadratic equations helps us to understand parabolas and their behavior, giving us crucial tools for a wide range of mathematical problems, from simple physics to economic modeling.
Calculator Approximations
Calculator approximations are essential for understanding and conveying solutions in practical terms. Often, exact expressions, such as \( \sqrt{14} \), may not be visually intuitive. Hence, we use calculators to convert these into decimal forms.
In this exercise, we approximate \( \sqrt{14} \) as \( 3.742 \), which is more straightforward to interpret and use in applications. Such approximations are crucial when we need to quickly assess the magnitude or when precision up to a certain number of decimal places is required, like in scientific research or engineering tasks.

Always round calculator approximations to the defined decimal places, ensuring consistency and accuracy in your computations. In this exercise, rounding to three decimal places offers a balance between precision and simplicity, providing a clear picture while remaining close to the exact value.

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

A sound level of \(\beta=120 \mathrm{db}\) is at the threshold of pain. (Some loud rock concerts reach this level.) The sound intensity that corresponds to \(\beta=120 \mathrm{db}\) is \(1 \mathrm{W} / \mathrm{m}^{2}\). Use this information and the equation \(\beta=10 \log _{10}\left(I / I_{0}\right)\) to determine \(I_{0}\), the intensity of a barely audible sound at the threshold of hearing. What is the decibel level, \(\beta\), of a barely audible sound?

Solve the equation \(\log _{2} x=\log _{x} 3 .\) For each root, give an exact expression and a calculator approximation rounded to two decimal places.

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$4\left(10-e^{x}\right) \leq-3$$

An article that appeared in the August \(13,1994,\) New York Times reported German authorities have discovered.... a tiny sample of weapons-grade nuclear material believed to have been smuggled out of Russia to interest foreign governments or terrorist groups that might want to build atomic bombs..... [the police] said they had seized the material, .028 ounces of highly enriched uranium- 235 in June in ... Bavaria ... and have since arrested... \([\text { six }]\) suspects. \(\ldots\) Suppose that the suspects, in an attempt to avoid arrest, had thrown the 0.028 ounces of uranium- 235 into the Danube River, where it would sink to the bottom. How many ounces of the uranium-235 would still be in the river after 1000 years? The half-life of uranium- 235 is \(7.1 \times 10^{8}\) years.

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\ln (2-5 x)>2$$

See all solutions

Recommended explanations on Math Textbooks

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.