Problem 23 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(1-\frac{1}{2 x+1}-\frac{1}{(2 x+1)^{2}}=0\)

Step-by-Step Solution

Answer

The equation has no solution.

Step 1: Identify and Isolate the Denominator

The given equation is \( 1 - \frac{1}{2x + 1} - \frac{1}{(2x + 1)^2} = 0 \). To solve this equation, first multiply through by \((2x + 1)^2 \), the least common denominator (LCD), to clear the fractions. Multiply each term in the equation by \((2x + 1)^2 \) and use the distributive property: \((2x + 1)^2 \times 1 - \frac{(2x + 1)^2}{2x + 1} - \frac{(2x + 1)^2}{(2x + 1)^2} = 0 \).

Step 2: Simplify the Equation

Simplify each term in the equation: \( (2x + 1)^2 - (2x + 1) - 1 = 0 \). Now the equation is a quadratic equation in terms of \(2x + 1 \).

Step 3: Expand and Rearrange

Expand \((2x + 1)^2 \): \((2x + 1)^2 = 4x^2 + 4x + 1 \). Substitute back into the simplified equation: \( 4x^2 + 4x + 1 - (2x + 1) - 1 = 0 \). Combine like terms: \( 4x^2 + 2x = 0 \).

Step 4: Factorize the Quadratic Equation

Factorize the quadratic equation: \( 4x^2 + 2x = 0 \). Factor out the greatest common factor (GCF): \( 2x (2x + 1) = 0 \).

Step 5: Solve for x

Set each factor equal to zero and solve: For \( 2x = 0 \): \( x = 0 \). For \( 2x + 1 = 0 \): \( x = - \frac{1}{2} \).

Step 6: Check the Solutions

Substitute \( x = 0 \) back into the original equation: \( 1 - \frac{1}{2(0) + 1} - \frac{1}{(2(0) + 1)^2} = 1 - 1 - 1 = -1 \), which is not true.This means that \( x = 0 \) is not a solution. Now substitute \( x = - \frac{1}{2} \) into the original equation: \( 1 - \frac{1}{2(- \frac{1}{2}) + 1} - \frac{1}{(2( -\frac{1}{2}) + 1)^2} = 1 - 1 - \frac{1}{0} \), which is not valid as it leads to a division by zero.This means that \( x = - \frac{1}{2} \) is not a solution either. Hence, the equation has no solution.

Key Concepts

Understanding the Least Common Denominator (LCD)

To solve the given equation, we need to clear the fractions by using the Least Common Denominator (LCD). The LCD is the smallest expression that can be evenly divided by each of the denominators in the equation. For instance, in the equation \[1 - \frac{1}{2x + 1} - \frac{1}{(2x + 1)^2} = 0\]the denominators are \[2x + 1 \text{ and } (2x + 1)^2\]. The LCD here is \[(2x + 1)^2\]. Multiplying every term in the equation by the LCD helps to eliminate the fractions and ease the solving process. Thus, our equation becomes: \[(2x + 1)^2 \cdot 1 - \frac{(2x + 1)^2}{2x+ 1} - \frac{(2x + 1)^2}{(2x+ 1)^2} = 0\]. This simplifies each term and results in a cleaner equation without fractions.

Simplifying Equations

Simplifying equations is about reducing the equation to its simplest form. Once we clear the denominators using the LCD, it's time to simplify each term. From our example, after multiplying by the LCD, we have: \[(2x + 1)^2 - (2x + 1) - 1 = 0\]. By simplifying, we eliminate complex fractions and reduce terms into basic polynomial or linear expressions. Simplifying makes the equation easier to handle, setting the stage for further operations such as factoring or solving for the variable.

Factoring Quadratic Equations

Factoring helps to break down a quadratic equation into simpler linear factors. After expanding and rearranging the simplified equation, we obtain: \[4x^2 + 2x = 0\]. To factorize, we look for common terms. Here, we can factor out \[2x\], leading us to: \[2x (2x + 1) = 0\]. Factoring helps to pinpoint the possible solutions (roots) of the equation by setting each factor equal to zero. So, we solve \[2x = 0\] and \[2x + 1 = 0\] to find the values of \[x\] that satisfy the original equation.

Checking Solutions

Once we find potential solutions, we must check them in the original equation. This step ensures that our solutions are valid. For instance, substituting \[x = 0\] into: \[1 - \frac{1}{2(0) + 1} - \frac{1}{(2(0) + 1)^2} = 0\]leads to \[-1\] which is false. Substituting \[x = - \frac{1}{2}\] results in a division by zero, which is not possible. Hence, both potential solutions fail the check, indicating there’s no valid solution for the equation. Always verify each root by plugging it back into the original equation to ensure it makes the equation true.