/*! 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 18 Solve each equation by the zero-... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 18

Solve each equation by the zero-factor property. $$-6 x^{2}+7 x=-10$$

Short Answer

Expert verified
\(\frac{2}{3}\) and \(\frac{5}{2}\)

Step by step solution

01

- Move all terms to one side

Add 10 to both sides of the equation to move all terms to one side: -6x^{2} + 7x + 10 = 0.
02

- Factor the quadratic equation

Factor the quadratic equation. In this case, we look for two numbers that multiply to -60 and add to 7. We can factor the equation as follows: -6x^{2} + 7x + 10 = 0 = (3x - 2)(-2x - 5) = 0
03

- Apply the zero-factor property

Set each factor to zero and solve for x : 3x - 2 = 0 2x - 5 = 0
04

- Solve for x

Solve the equations from the previous step:3x - 2 = 0 :\[3x - 2 = 0\]\[3x = 2 \] \[ x =\frac{2}{3} \]5x - 2 = 0 : \[2x - 5 = 0\] \[2x = 5\] \[x = \frac{5}{2}\]

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.

solving quadratic equations
Quadratic equations are equations that involve the variable being squared. They follow the general format: ax^2 + bx + c = 0. Here, 'a', 'b', and 'c' are constants. To solve these equations, we aim to find the value(s) of 'x' that make the equation true. There are several methods to solve such equations, like the quadratic formula, completing the square, and factoring. We often use the zero-factor property through factoring when it is possible. This provides an algebraic way to find solutions to the equation by breaking it down into simpler factors.
factoring quadratic expressions
Factoring quadratic expressions is a method where we rewrite the quadratic equation as a product of two binomials. For example, a quadratic of the form: ax^2 + bx + c can sometimes be expressed as (dx + e)(fx + g). To factor successfully, we typically follow these steps:
  • Identify coefficients 'a' (leading coefficient), 'b' (linear coefficient), and 'c' (constant term).
  • Find two numbers that multiply to 'ac' and add to 'b'.
  • Use these numbers to split the middle term and factor by grouping.
  • Check the factors by multiplying them to ensure they equal the original quadratic expression.
In our example, we factored -6x^2 + 7x + 10 into (3x - 2)(-2x - 5).
zero product principle
The zero product principle states that if you have a product of factors that equals zero, at least one of the factors must be zero. In algebra, this principle is very useful for solving equations. For instance, if we have an equation like (a)(b) = 0, we know that either a = 0, b = 0, or both. This allows us to set each factor equal to zero and solve for the variable separately. Going back to our example, we set (3x - 2) = 0 and (2x - 5) = 0, which simplifies our work by splitting the problem into two smaller and more manageable pieces.
algebraic solutions
Algebraic solutions involve finding the exact values of the variable that satisfy the equation. Once we factor our quadratic and apply the zero product principle, we solve for 'x' algebraically. Using our previous steps, we've transformed -6x^2 + 7x + 10 into (3x - 2)(-2x - 5) = 0. By solving 3x - 2 = 0, we get x = 2/3, and by solving 2x - 5 = 0, we get x = 5/2. These individual solutions represent the points where the original quadratic equation equals zero. This method is precise and reliable for finding exact solutions, essential in many applications of quadratic equations.

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

Simplify each power of i. $$i^{-13}$$

Classroom Ventilation According to the American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (ASHRAE), a nonsmoking classroom should have a ventilation rate of \(15 \mathrm{ft}^{3}\) per min for each person in the room. (a) Write an equation that models the total ventilation \(V\) (in cubic feet per hour) necessary for a classroom with \(x\) students. (b) A common unit of ventilation is air change per hour (ach). 1 ach is equivalent to exchanging all of the air in a room every hour. If \(x\) students are in a classroom having volume \(15,000 \mathrm{ft}^{3},\) determine how many air exchanges per hour \((A)\) are necessary to keep the room properly ventilated. (c) Find the necessary number of ach ( \(A\) ) if the classroom has 40 students in it. (d) In areas like bars and lounges that allow smoking, the ventilation rate should be increased to \(50 \mathrm{ft}^{3}\) per min per person. Compared to classrooms, ventilation should be increased by what factor in heavy smoking areas?

Solve each equation or inequality. $$|3 x+2| \leq 0$$

Find each product. Write the answer in standard form. $$(\sqrt{6}+i)(\sqrt{6}-i)$$

If p units of an item are sold for \(x\) dollars per unit, the revenue is \(R=p x\). Use this idea to analyze the following problem. Number of Apartments Rented The manager of an 80-unit apartment complex knows from experience that at a rent of \(\$ 300,\) all the units will be full. On the average, one additional unit will remain vacant for each \(\$ 20\) increase in rent over \(\$ 300 .\) Furthermore, the manager must keep at least 30 units rented due to other financial considerations. Currently, the revenue from the complex is \(\$ 35,000 .\) How many apartments are rented? According to the problem, the revenue currently generated is \(\$ 35,000 .\) Substitute this value for revenue into the equation from Exercise \(53 .\) Solve for \(x\) to answer the question in the problem.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and 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.