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Each exercise contains an error. Find and correct the error. See the Concept Check in the section. $$\begin{aligned} &(x-5)(x+2)=0\\\ &x-5=0 \text { or } x+2=0\\\ &x=-5 \quad \text { or } \quad x=-2 \end{aligned}$$

Short Answer

Expert verified
The error was in solving the equations. Correct solutions: \(x = 5\) or \(x = -2\).

Step by step solution

01

Identify the Error

The error lies in the solutions for the equations set equal to zero. After factoring, the zero-product property should be used correctly to find the values of \(x\).
02

First Equation Setup

Given \((x-5)(x+2) = 0\), apply the zero-product property to set each factor to zero separately: \(x-5 = 0\) or \(x+2 = 0\).
03

Solve First Equation

Solve \(x-5=0\) by adding 5 to both sides: \(x = 5\).
04

Solve Second Equation

Solve \(x+2=0\) by subtracting 2 from both sides: \(x = -2\).
05

Conclusion

The correct solutions to the equation \((x-5)(x+2) = 0\) are \(x = 5\) and \(x = -2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring
Factoring is a fundamental concept in algebra that involves breaking down a complex expression into simpler components, or "factors," that, when multiplied, give back the original expression. Understanding factoring is crucial in solving equations, especially quadratic equations. In this context, the expression \((x-5)(x+2)\) is already factored because it is expressed as a product of two binomials.

The goal of factoring is often to simplify expressions, making them easier to work with in solving equations. Later, these factors can be used in conjunction with the zero-product property, a key principle in solving polynomial equations. This property states that if a product of factors equals zero, then at least one of the factors must be zero.

To factor a quadratic polynomial, one usually looks for two numbers that multiply to give the constant term and add to give the coefficient of the middle term (in the case of quadratic trinomials). However, in the given exercise, the polynomial is already provided in its factored form, one step ahead in solving it.
Solving Quadratic Equations
When solving quadratic equations like \((x-5)(x+2) = 0\), the most efficient method is to use the zero-product property. This principle simplifies the solution process by allowing you to set each factor equal to zero separately, since if the product of any two numbers is zero, one of them must be zero.

Here's how it works in practice:
  • Start by setting each factor equal to zero, giving two separate equations: \(x-5 = 0\) and \(x+2 = 0\).
  • Solve these simple linear equations individually.
In this example:
  • From \(x-5 = 0\), adding 5 to both sides gives \(x = 5\).
  • From \(x+2 = 0\), subtracting 2 from both sides gives \(x = -2\).
It's important to solve each resulting equation carefully to ensure accuracy in finding all possible solutions to the original problem.
Error Detection in Algebra
Error detection is an essential skill in algebraic problems solving. Even slight mistakes can lead to incorrect solutions, as seen in the original solution attempt. Initially, the solutions listed were \(x = -5\) or \(x = -2\), which were incorrect due to a simple algebraic oversight.

To effectively detect errors:
  • Carefully check each step of your solution process.
  • Verify calculations and ensure all operations follow algebraic rules.
  • Review each factor and equation setup to catch any discrepancies early on.
In the given exercise, applying the zero-product property correctly ensures that each factor accounts for a potential solution by setting them independently to zero. By practicing meticulous error checking, students can avoid common pitfalls and deepen their understanding of solving algebraic equations.

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Most popular questions from this chapter

Find the value of \(c\) that makes each trinomial a perfect square trinomial. Factor \(x^{6}-1\) completely, using the following methods from this chapter. A. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) B. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) C. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\)

Factor. Assume that variables used as exponents represent positive integers. $$ 9 x^{2 n}+24 x^{n}+16 $$

The function \(f(x)=0.0007 x^{2}+0.24 x+7.98\) can be used to approximate the total cheese production in the United States from 2000 to \(2009,\) where \(x\) is the number of years after 2000 and \(y\) is pounds of cheese (in billions). Round answers to the nearest hundredth of a billion. (Source: National Agricultural Statistics Service, USDA) a. Approximate the number of pounds of cheese produced in the United States in 2000. b. Approximate the number of pounds of cheese produced in the United States in 2005. c. Use this function to estimate the pounds of cheese produced in the United States in 2015. d. From parts \((a),(b),\) and \((c),\) determine whether the number of pounds of cheese produced in the United States is increasing at a steady rate. Explain why or why not.

Factor. Assume that variables used as exponents represent positive integers. $$ 25 x^{2 n}-81 $$

Explain why \((-5)^{0}\) simplifies to 1 but \(-5^{0}\) simplifies to \(-1\)

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