/*! 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 94 Solve each equation. $$ (5 x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 94

Solve each equation. $$ (5 x+6)^{2}=0 $$

Short Answer

Expert verified
x = -\frac{6}{5}

Step by step solution

01

Understand the Equation

The given equation is \( (5x + 6)^{2} = 0 \). This is a quadratic equation set to zero, which suggests the square of a binomial is zero.
02

Apply the Zero-Product Property

Recall that if \(a^{2} = 0\), then \(a = 0\). Hence, if \((5x + 6)^{2} = 0\), then \(5x + 6 = 0\).
03

Solve for x

To find the value of x, isolate x in the equation \(5x + 6 = 0\). First, subtract 6 from both sides to get \(5x = -6\). Then, divide both sides by 5 to get \(\frac{5x}{5} = \frac{-6}{5}\), which simplifies to \(x = -\frac{6}{5}\).
04

Verification

To verify, substitute \(x = -\frac{6}{5}\) back into the original equation: \((5(-\frac{6}{5}) + 6)^{2} = 0\). This simplifies to \(( - 6 + 6)^{2} = 0\), or \(0^{2} = 0\), which is true. Hence, \(x = -\frac{6}{5}\) is correct.

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 Equation
A quadratic equation is any equation that can be rearranged into the form \[ax^2 + bx + c = 0\] where 'a', 'b', and 'c' are constants, and 'x' represents an unknown variable. In a quadratic equation, the highest power of the variable 'x' is 2. Quadratic equations typically have two solutions, which can be found using the quadratic formula, factoring, or other methods.

In our exercise, we have the equation \[(5x + 6)^2 = 0\]. Since this is in the form of a squared binomial set to zero, it still falls under the category of a quadratic equation.
Zero-Product Property
The zero-product property is a crucial principle in solving quadratic equations. It states that if the product of two factors is zero, then at least one of the factors must be zero. Mathematically, it can be written as: if \[ab = 0\], then \[a = 0\] or \[b = 0\].
In the given problem, our equation was \[(5x + 6)^2 = 0\]. Using the zero-product property, we can infer that if a square of any number equals zero, then the number itself must be zero. Therefore, we set \[5x + 6 = 0\] to find the solution.
Binomial
A binomial is an algebraic expression that contains exactly two terms. It is of the form \[a + b\] or \[a - b\], where 'a' and 'b' are terms which can be numbers, variables, or products of both.

In our problem, the expression \[5x + 6\] is a binomial. When we squared this binomial in the equation \[(5x + 6)^2 = 0\], we dealt specifically with the binomial form. Squaring a binomial follows the formula \[(a + b)^2 = a^2 + 2ab + b^2\], but in this exercise, it is more straightforward since we are setting the squared binomial to zero directly.

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 formula for calculating the distance \(d\) one can see from an airplane to the horizon on a clear day is $$ d=1.22 \sqrt{x} $$ where \(x\) is the altitude of the plane in feet and \(d\) is given in miles. How far can one see to the horizon in a plane flying at the following altitudes? (a) \(15,000\) ft (b) \(18,000\) ft (c) \(24,000 \mathrm{ft}\)

Simplify each radical. Assume that all variables represent nonnegative real numbers. $$ \sqrt{m^{2}} $$

Solve each equation. (Hint: In Exercises 67 and 68, extend the concepts to fourth root radicals.) $$ \sqrt[3]{2 x}=\sqrt[3]{5 x+2} $$

Simplify each radical. Assume that all variables represent nonnegative real numbers. $$ \sqrt{p^{17}} $$

Simplify each radical. $$ \sqrt[3]{216 m^{3} n^{6}} $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Mechanics 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.