/*! 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 69 Find the real solutions of each ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 69

Find the real solutions of each equation. $$ 3 x^{-2}-7 x^{-1}-6=0 $$

Short Answer

Expert verified
The solutions are \( x = -\frac{3}{2} \) and \( x = \frac{1}{3} \).

Step by step solution

01

Rewrite the Equation

Rewrite the given equation by substituting a new variable. Let \( y = x^{-1} \). The equation becomes: \[ 3 y^2 - 7 y - 6 = 0 \]
02

Factor the Quadratic Equation

Now, factor the quadratic equation \( 3 y^2 - 7 y - 6 = 0 \). The factors are: \[ (3y + 2)(y - 3) = 0 \]
03

Solve for y

Solve for the variable \( y \) using the factored equation: \[ 3y + 2 = 0 \implies y = -\frac{2}{3} \] \[ y - 3 = 0 \implies y = 3 \]
04

Substitute Back the Original Variable

Replace \( y \) with \( x^{-1} \). We get two equations to solve for \( x \): \[ x^{-1} = -\frac{2}{3} \implies x = -\frac{3}{2} \] \[ x^{-1} = 3 \implies x = \frac{1}{3} \]
05

Check for Real Solutions

Verify that the solutions we found are real numbers, which they are. Therefore, the real solutions to the equation are: \( x = -\frac{3}{2} \) and \( x = \frac{1}{3} \)

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.

quadratic equations
A quadratic equation is a second-order polynomial equation in a single variable. It typically follows the standard form:






cannot be solved easily using simple algebraic methods. To find the roots or solutions of a quadratic equation, we often use methods like factoring, completing the square, or applying the quadratic formula:
The quadratic formula is especially useful when the equation cannot be factored easily. It's given by: In the context of the given exercise, rewriting the equation using the new variable converts the problem into a quadratic equation. The next step involves solving this quadratic equation to find the solutions.
substitution method
The substitution method is a powerful technique used to simplify complex equations. In this exercise, we started with where is a new variable representing. By making this substitution, we transformed the original equation into a simpler quadratic form:
  • This makes it easier to handle since quadratic equations have well-established methods for finding solutions.
  • After solving the quadratic equation for we substitute back to find the values of the original variable

For example, in the step-by-step solution:
  • We let
  • The equation becomes
  • Once we find we revert to find by solving equations
This technique simplifies the problem by breaking it into smaller, manageable parts.
factoring polynomials
Factoring is one of the primary methods for solving quadratic equations. It involves expressing the quadratic polynomial as a product of two binomials. The factored form of a quadratic polynomial helps us identify the roots easily. Consider the quadratic equation from the exercise:
  • First, we factorize it as
  • This factorization allows us to solve the equation by setting each factor equal to zero:
  • Solving these simple equations gives the values of
is a widely used technique, especially when the polynomial can be easily decomposed into a product of simpler expressions.
inverse functions
Inverse functions reverse the effect of the original function. If a function

  • For the given exercise, after substitution and solving for we need to go back and find the values of Here, functions as an inverse function. When we find
  • This involves reversing the change made by substituting, essentially applying the inverse function. In this case:
  • 路聽Solving gives us the real solutions:
  • The notion of inverse functions is critical in solving equations where a variable substitution has been used. It ensures that we find the correct solutions to the original problem.

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 the real solutions, if any, of each equation. Use any method. $$ 2+z=6 z^{2} $$

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

Geometry Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

Solve: \(\sqrt{3 x+5}-\sqrt{x-2}=\sqrt{x+3}\)

Find \(k\) so that the equation \(x^{2}-k x+4=0\) has a repeated real solution.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.