/*! 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 80 Solve: \((x-1)^{2}=5\) (Sectio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 80

Solve: \((x-1)^{2}=5\) (Section \(1.5, \text { Example } 2)\)

Short Answer

Expert verified
The solution of the equation \((x-1)^{2}=5\) are \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\).

Step by step solution

01

Apply the square root to both sides

Take the square root of the both sides of the equation \((x-1)^{2}=5\), these results in two equations: \(x-1= \sqrt{5}\) and \(x-1= -\sqrt{5}\) because \(\sqrt{a^2}\) can be either \(a\) or \(-a\).
02

Solve for x

Solving the above equations for x gives \(x=1+\sqrt{5}\) for the first equation and \(x=1-\sqrt{5}\) for the second equation.
03

Verify the solution

Substitute \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\) into the original equation. For \(x=1+\sqrt{5}\), you get \((1+\sqrt{5}-1)^{2}=(\sqrt{5})^{2}=5\), which matches the original equation. For \(x=1-\sqrt{5}\), you get \((1-\sqrt{5}-1)^{2}=(-\sqrt{5})^{2}=5\), which also matches the original equation. Therefore, the solutions are verified.

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.

Square Root Property
The square root property is a powerful tool in solving quadratic equations. It states that if you have an equation of the form \((x - h)^2 = k\), then taking the square root of both sides gives two possible equations:
  • \(x - h = \sqrt{k}\)
  • \(x - h = -\sqrt{k}\)
This duality occurs because both a number and its negative have the same square.
In the original exercise with \((x-1)^2=5\), we apply this property and get \(x-1 = \sqrt{5}\) and \(x-1 = -\sqrt{5}\). This results in two possible values for \(x\) once we solve both equations.
Using the square root property effectively requires recognizing when an equation is in the right format and thinking about the implications of taking a square root.
Solving Equations
Solving equations involves finding values for the variable that make the equation true. Once you've applied the square root property and arrived at the equations \(x-1=\sqrt{5}\) and \(x-1=-\sqrt{5}\), the next step is straightforward. You want to solve for \(x\) by isolating it on one side of the equation.
  • Start by adding 1 to both sides to get \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\).
  • These solutions represent the values of \(x\) that will satisfy the original equation.
These steps test our ability to perform basic algebraic manipulations.
The simplicity of this problem helps build understanding of more complex algebraic concepts over time.
Verification of Solutions
Verification of solutions is a vital step in solving equations. It ensures that the operations performed have led to valid solutions. By substituting the solutions back into the original equation, you verify their correctness:
  • For \(x=1+\sqrt{5}\), substitute back to check \([1+\sqrt{5} - 1]^2 = (\sqrt{5})^2 = 5\).
  • For \(x=1-\sqrt{5}\), verify similarly: \([1-\sqrt{5} - 1]^2 = (-\sqrt{5})^2 = 5\).
Both instances result in true statements, confirming that the solutions satisfy the original equation.
This process is crucial not just for accuracy, but it also builds confidence in solving equations and reinforces understanding of algebraic principles.

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

Exercises \(127-129\) will help you prepare for the material covered in the next section. Determine the amplitude and period of \(y=10 \cos \frac{\pi}{6} x\).

This exercise is intended to provide some fun with biorhythms, regardless of whether you believe they have any validity. We will use each member鈥檚 chart to determine biorhythmic compatibility. Before meeting, each group member should go online and obtain his or her biorhythm chart. The date of the group meeting is the date on which your chart should begin. Include 12 months in the plot. At the meeting, compare differences and similarities among the intellectual sinusoidal curves. Using these comparisons, each person should find the one other person with whom he or she would be most intellectually compatible.

Use a graphing utility to graph two periods of the function. $$y=0.2 \sin \left(\frac{\pi}{10} x+\pi\right)$$

In Exercises \(115-116,\) convert each angle to \(D^{\circ} M^{\prime} S^{\prime \prime}\) form. Round your answer to the nearest second. $$ 30.42^{\circ} $$

Assuming Earth to be a sphere of radius 4000 miles, how many miles north of the Equator is Miami, Florida, if it is \(26^{\circ}\) north from the Equator? Round your answer to the nearest mile.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

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.