/*! 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 each equation. (All soluti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 80

Solve each equation. (All solutions for these equations are nonreal complex numbers.) \((x-5)^{2}=-36\)

Short Answer

Expert verified
x = 5 卤 6i

Step by step solution

01

- Recognize the form of the equation

The equation given is \( (x-5)^{2}=-36 \). Notice that it is set equal to a negative number, which means the solutions will be complex numbers.
02

- Apply the square root

Take the square root of both sides of the equation to isolate \( x-5 \): \[ \sqrt{(x-5)^{2}} = \sqrt{-36} \]. Recall that the square root of a square is the absolute value of the interior term. Therefore, this simplifies to: \[ x-5 = \pm \sqrt{-36} \].
03

- Simplify the square root of a negative number

Simplify the square root of \( -36 \) using the property \( \sqrt{-a} = i\sqrt{a} \), where \( i \) is the imaginary unit: \[ x-5 = \pm i \sqrt{36} = \pm 6i \].
04

- Solve for x

Solve for \( x \) by adding \( 5 \) to both sides of the equation: \[ x = 5 \pm 6i \].

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.

complex numbers
A complex number is a number that has both a real part and an imaginary part. The standard form for a complex number is written as: \(a + bi\), where:
  • \(a\) is the real part
  • \(bi\) is the imaginary part
The imaginary unit \(i\) is defined as \(\sqrt{-1}\). This definition leads to the fact that when you square \(i\), you get -1: \(i^2 = -1\). Complex numbers are essential in solving equations that don't have real solutions. These numbers extend our number system and are very useful in various fields like engineering and physics. In our exercise, the solution \(5 \pm 6i\) includes both real and imaginary parts.
square root property
The square root property is used to solve equations of the form \((x-a)^2 = b\). The steps are:
  • Take the square root of both sides of the equation.
  • Simplify the resulting expressions.
  • Don't forget the \(\pm\) symbol when taking the square root, as there are two possible solutions.
In our example, we start with \((x-5)^2 = -36\). By taking the square root of both sides, we get \(x - 5 = \pm \sqrt{-36}\). This is a crucial step in isolating the variable and finding our solutions.
imaginary unit
The imaginary unit, denoted as \(i\), is defined as \(\sqrt{-1}\). This unit is essential when dealing with the square roots of negative numbers. Key properties of the imaginary unit include:
  • \(i^2 = -1\)
  • \(\sqrt{-a} = i\sqrt{a}\), for any positive real number \(a\)
In our exercise, this property allows us to simplify \(\sqrt{-36}\) as \(6i\). This transformation helps us solve the equation effectively, even when dealing with negative numbers under the square root.
isolating variables
Isolating the variable is a crucial step in solving any algebraic equation. It involves manipulating the equation to get the variable by itself on one side of the equation. The process typically includes:
  • Using addition or subtraction to move terms from one side to the other.
  • Using multiplication or division to simplify the coefficients of the variable.
In our exercise, after simplifying the square root, we had \(x - 5 = \pm 6i\). We then added 5 to both sides to finally isolate \(x\): \(x = 5 \pm 6i\).

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 ball is projected upward from ground level, and its distance in feet from the ground in t seconds is given by $$ s(t)=-16 t^{2}+160 t $$ After how many seconds does the ball reach a height of \(400 \mathrm{ft}\) ? Describe in words its position at this height.

The following exercises are not grouped by type. Solve each equation. $$6=7(2 w-3)^{-1}+3(2 w-3)^{-2}$$

Solve each equation. Check the solutions. $$(x-4)^{2}+(x-4)-20=0$$

Solve each problem using a quadratic equation. Use the formula \(A=P(1+r)^{2}\) to find the interest rate \(r\) at which a principal \(P\) of \(\$ 10,000\) will increase to \(\$ 10,920.25\) in 2 yr.

The following exercises are not grouped by type. Solve each equation. $$8 x^{6}+513 x^{3}+64=0$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

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.