/*! 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 70 Solve each equation in the compl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 70

Solve each equation in the complex number system. $$ 13 x^{2}+1=6 x $$

Short Answer

Expert verified
The solutions are \( x = \frac{3}{13} + \frac{2i}{13} \) and \( x = \frac{3}{13} - \frac{2i}{13} \).

Step by step solution

01

- Rearrange the Equation

First, rearrange the equation to set it to zero form: \[ 13x^2 + 1 = 6x \] becomes \[ 13x^2 - 6x + 1 = 0 \]
02

- Identify a, b, and c

Identify the coefficients from the quadratic equation \(ax^2 + bx + c = 0\). Here, \(a = 13\), \(b = -6\), and \(c = 1\).
03

- Apply the Quadratic Formula

Use the quadratic formula to find the solutions: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting in the values, we get: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 13 \cdot 1}}{2 \cdot 13} \] becomes \[ x = \frac{6 \pm \sqrt{36 - 52}}{26} \] which simplifies to \[ x = \frac{6 \pm \sqrt{-16}}{26} \]
04

- Simplify the Expression

Simplify the square root term: \[ \sqrt{-16} = 4i \] The expression becomes: \[ x = \frac{6 \pm 4i}{26} \]
05

- Divide Each Term

Divide each term in the numerator by the denominator: \[ x = \frac{6}{26} \pm \frac{4i}{26} \] which simplifies to \[ x = \frac{3}{13} \pm \frac{2i}{13} \]

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
In algebra, you might come across equations that don't have any real-number solutions. This is where complex numbers become handy. Complex numbers are numbers that have both a real and an imaginary part. They look like this:

\( a + bi \)

Here, \( a \) is the real part, and \( bi \) is the imaginary part. The letter \( i \) represents the imaginary unit, which is defined as \( i = \sqrt{-1} \). In our equation, we got solutions that involve an imaginary number because the square root of a negative number, like \( \-16 \), leads to an imaginary result. This allows us to extend the reach of quadratic equations beyond real numbers.
quadratic formula
The quadratic formula is a critical tool for solving equations of the form \( ax^2 + bx + c = 0 \). It's given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

To use it, you need to identify the coefficients \( a \), \( b \), and \( c \) from your equation. In our problem, \( a = 13 \), \( b = -6 \), and \( c = 1 \). Be sure to be careful with the signs! Then, substitute these values into the quadratic formula. The part under the square root, \( b^2 - 4ac \), is crucial. If it's negative, like in our example, you'll end up with an imaginary number. The quadratic formula helps you find both the real and complex solutions for any quadratic equation.
imaginary numbers
Imaginary numbers sound tricky but are essential for solving some equations. They come into play when you need the square root of a negative number. The standard imaginary unit is represented by \( i \), where \( i = \sqrt{-1} \).

In our solution, we came across \( \sqrt{-16} \). Simplifying this involves recognizing that:

\( \sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i \)

Here, \( 4 \) is the normal square root part, and \( i \) accounts for the negative under the square root. By using imaginary numbers, we can express any solution, even when it involves square roots of negative numbers, and handle them just like other numbers when performing operations.

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

Find all the real solutions of \(12 x^{7 / 5}+3 x^{2 / 5}=13 x^{9 / 10}\).

Reducing the Size of a Candy Bar A jumbo chocolate bar with a rectangular shape measures 12 centimeters in length, 7 centimeters in width, and 3 centimeters in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by \(10 \%\). To accomplish this reduction, management decides that the new bar should have the same 3 -centimeter thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar?

On a recent flight from Phoenix to Kansas City, a distance of 919 nautical miles, the plane arrived 20 minutes early. On leaving the aircraft, I asked the captain, "What was our tail wind?" He replied,"I don't know, but our ground speed was 550 knots." Has enough information been provided for you to find the tail wind? If possible, find the tail wind. \((1 \mathrm{knot}=1\) nautical mile per hour)

Challenge Problem If \(\frac{2}{3} \leq \frac{5-x}{3} \leq 3,\) find the largest possible value of \(2 x^{2}-3\)

The purity of gold is measured in karats, with pure gold being 24 karats. Other purities of gold are expressed as proportional parts of pure gold. Thus, 18-karat gold is \(\frac{18}{24},\) or \(75 \%\) pure gold; 12 -karat gold is \(\frac{12}{24},\) or \(50 \%\) pure gold; and so on. How much 12-karat gold should be mixed with pure gold to obtain 60 grams of 16-karat gold?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

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.