/*! 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 99 Find all real solutions of the e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 99

Find all real solutions of the equation. $$2 x+\sqrt{x+1}=8$$

Short Answer

Expert verified
The real solution is \(x = 3\).

Step by step solution

01

Isolate the Square Root

To simplify the problem, let's first isolate the square root term. We begin by moving the linear term, \(2x\), to the other side of the equation:\[\sqrt{x+1} = 8 - 2x.\]
02

Square Both Sides

Next, eliminate the square root by squaring both sides of the equation. Be careful to square the entire right side as a whole:\[(\sqrt{x+1})^2 = (8 - 2x)^2.\]This simplifies to:\[x+1 = (8-2x)^2 = 64 - 32x + 4x^2.\]
03

Rearrange into a Quadratic Equation

Now, rearrange the terms to form a quadratic equation:\[4x^2 - 32x + 64 - x - 1 = 0.\]Simplifying gives:\[4x^2 - 33x + 63 = 0.\]
04

Solve the Quadratic Equation

Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -33\), and \(c = 63\), to find the solutions:\[x = \frac{-(-33) \pm \sqrt{(-33)^2 - 4 \cdot 4 \cdot 63}}{2 \cdot 4}.\]Calculate:\[x = \frac{33 \pm \sqrt{1089 - 1008}}{8} = \frac{33 \pm \sqrt{81}}{8} = \frac{33 \pm 9}{8}.\]Thus, the solutions are:\[x = \frac{42}{8} = 5.25 \quad \text{and} \quad x = \frac{24}{8} = 3.\]
05

Verify the Solutions

We must verify the potential solutions in the original equation to check for extraneous solutions. Starting with \(x = 3\):\[2(3) + \sqrt{3+1} = 6 + 2 = 8.\]This satisfies the original equation.Now for \(x = 5.25\):\[2(5.25) + \sqrt{5.25+1} = 10.5 + \sqrt{6.25} = 10.5 + 2.5 = 13.\]\(x = 5.25\) does not satisfy the original equation, hence it is extraneous.

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.

Radical Equations
Radical equations are those that contain a variable within a root, typically a square root. To solve these types of equations, it's important to isolate the radical on one side before proceeding with further steps, such as squaring both sides.
Squaring both sides is a common technique to eliminate the square root, but it should be done carefully. This process can often introduce extraneous roots, solutions that appear valid but do not actually satisfy the original equation.
It's important to recheck potential solutions back in the original equation to ensure they work, which helps identify and exclude any extraneous solutions.
Quadratic Formula
The quadratic formula is a universal method for finding the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
This formula is particularly useful when factoring the quadratic is not straightforward. It provides two solutions: one using the plus sign and one using the minus sign in front of the square root.
When using the quadratic formula, it's crucial to substitute the correct values for \(a\), \(b\), and \(c\), then simplify carefully to reach the two potential solutions. These solutions may still require verification to determine if they fit within the context of the original equation.
Extraneous Solutions
Extraneous solutions are extra answers that arise from the processes used to solve an equation, but do not actually satisfy the original problem. They often occur when both sides of an equation are squared, as this can introduce additional roots.
It's crucial to identify extraneous solutions by substituting each solution back into the original equation. Only those values that satisfy the equation should be considered valid solutions.
To manage extraneous solutions effectively, always keep the original context of the equation in mind and verify potential solutions thoroughly. This precautionary step ensures that you report only the true solutions to the problem at hand.

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

Complete the following tables. What happens to the \(n\) th root of 2 as \(n\) gets large? What about the \(n\) th root of \(\frac{1}{2} ?\) $$\begin{array}{|r|r|}\hline n & 2^{1 / n} \\\\\hline 1 & \\\2 & \\\5 & \\\10 & \\\100 & \\\\\hline\end{array}$$ $$\begin{array}{|r|r|} \hline n & \left(\frac{1}{2}\right)^{1 / n} \\\\\hline 1 & \\\2 & \\\5 & \\\10 & \\\100 & \\\\\hline\end{array}$$ Construct a similar table for \(n^{1 / n}\). What happens to the \(n\) th root of \(n\) as \(n\) gets large?

Solve the equation for the variable \(x\). The constants \(a\) and \(b\) represent positive real numbers. $$\sqrt{x}-a \sqrt[3]{x}+b \sqrt[6]{x}-a b=0$$

Simplify the expression. (a) \(\sqrt{81 x^{2}+81}\) (b) \(\sqrt{36 x^{2}+36 y^{2}}\)

A determined gardener has \(120 \mathrm{ft}\) of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area that is enclosed to be at least \(800 \mathrm{ft}^{2}\). What range of values is possible for the length of her garden?

Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force \(F\) acting on an object situated on this line segment is $$F=\frac{-K}{x^{2}}+\frac{0.012 K}{(239-x)^{2}}$$ where \(K>0\) is a constant and \(x\) is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the "dead spot" where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.) PICTURE CANT COPY
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

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.