Problem 4
Question
Find and correct the error at the right. \(\begin{aligned}(2 x+4)(x-2) &=0 \\ 2 x+4 &=0 \quad \text { or } \quad x-2=0 \\\ 2 x &=4 \\ x &=2 \end{aligned}\)
Step-by-Step Solution
Verified Answer
The correct solutions for the equation are \( x = -2 \) and \( x = 2 \). The error in the original exercise occurred during simplification of the first part (factor), resulting in an incorrect solution for x.
1Step 1: Equation Overview
The equation given is \( (2 x+4)(x-2) = 0 \). To solve this equation, each of the factors need to be set equal to zero and solved for x.
2Step 2: Identify and Correct the Error
The next step is \( 2x + 4 = 0 \) or \( x - 2 = 0 \). When the first part is simplified, it results in \( 2x = 4 \), which gives \( x = 2 \). However, the actual simplification should be: Subtracting 4 from both sides gets \( 2x = -4 \), then dividing both sides by 2 gives \( x = -2 \). The second part \( x - 2 = 0 \) is solved correctly, with \( x = 2 \).
3Step 3: Statement of Correct Solutions
The correct solutions are \( x = -2 \) or \( x = 2 \).
Key Concepts
FactoringZero Product PropertySolving Equations
Factoring
Factoring a polynomial involves breaking it down into simpler expressions that multiply together to give the original polynomial.
This is especially useful when solving quadratic equations. In this exercise, we are given the equation \((2x+4)(x-2) = 0\). It is already factored, which makes application of the Zero Product Property straightforward.
When factoring, look for common factors that can be factored out of each term. If the polynomial is not already factored, you would often start by factoring out the greatest common factor (GCF).
In some cases, you might use techniques like splitting the middle term or applying special product formulas, such as the difference of squares. This factorization helps in setting up the equation for the next step of solving using the Zero Product Property.
This is especially useful when solving quadratic equations. In this exercise, we are given the equation \((2x+4)(x-2) = 0\). It is already factored, which makes application of the Zero Product Property straightforward.
When factoring, look for common factors that can be factored out of each term. If the polynomial is not already factored, you would often start by factoring out the greatest common factor (GCF).
In some cases, you might use techniques like splitting the middle term or applying special product formulas, such as the difference of squares. This factorization helps in setting up the equation for the next step of solving using the Zero Product Property.
Zero Product Property
The Zero Product Property is a crucial concept in solving quadratic equations that have been factored.
It states that if the product of two expressions is zero, then at least one of the expressions must be zero: \(A \times B = 0\) means either \(A = 0\) or \(B = 0\).
This property simplifies the process of solving equations, as shown with the equation \((2x+4)(x-2) = 0\).
To apply the Zero Product Property, set each factor equal to zero:
It states that if the product of two expressions is zero, then at least one of the expressions must be zero: \(A \times B = 0\) means either \(A = 0\) or \(B = 0\).
This property simplifies the process of solving equations, as shown with the equation \((2x+4)(x-2) = 0\).
To apply the Zero Product Property, set each factor equal to zero:
- \(2x + 4 = 0\)
- \(x - 2 = 0\)
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true.
For quadratic equations, this often means applying methods like factoring and using the Zero Product Property.
In our exercise, after factoring the equation and using the Zero Product Property, we end up with two simple equations:
Practice with multiple problems to build confidence in each step, ensuring accuracy in finding solutions.
For quadratic equations, this often means applying methods like factoring and using the Zero Product Property.
In our exercise, after factoring the equation and using the Zero Product Property, we end up with two simple equations:
- \(2x + 4 = 0\)
- \(x - 2 = 0\)
- Subtract 4 from both sides: \(2x = -4\)
- Divide both sides by 2: \(x = -2\)
- \(x = 2\)
Practice with multiple problems to build confidence in each step, ensuring accuracy in finding solutions.
Other exercises in this chapter
Problem 4
Match the trinomial with a correct factorization. $$ \begin{aligned} &A.)\quad (x+5)(x-4)\\\ &B.)\quad(x+4)(x+5)\\\ &C.)\quad(x-4)(x-5)\\\ &D.)\quad(x+4)(x-5) \
View solution Problem 4
Use a special product pattern to find the product. $$ (w+11)(w-11) $$
View solution Problem 4
$$ (3 x+4)(2 x-1)=3 x(?)+4(?) $$
View solution Problem 5
Identify the polynomial by degree and by the number of terms. $$ 12 x^{2}+7 x $$
View solution