Problem 16 - Elementary Algebra

Question

For the following problems, solve the equations, if possible. $$ (2 x+1)(x-7)=0 $$

Step-by-Step Solution

Answer

Based on the given step-by-step solution, the short answer below summarizes the process to solve the quadratic equation \((2x+1)(x-7)=0\): The quadratic equation \((2x+1)(x-7)=0\) is given in factored form. To find the solution, each factor is individually set equal to zero and solved for \(x\). Setting \(2x + 1 = 0\) and simplifying, we find that \(x = -1/2\). Setting \(x - 7 = 0\), we find that \(x = 7\). Therefore, the solutions to the equation are \(x = -1/2\) and \(x = 7\). Both of these values make the original equation true.

Step 1: Set each factor equal to zero

Since the product of two factors is zero, at least one of those factors must be zero. We will set each factor equal to zero and solve for \(x\) in each case. $$ (2x + 1) = 0 \\ (x - 7) = 0 $$

Step 2: Solve each equation

Now we will solve both equations for \(x\) For the first equation \((2x + 1) = 0\), subtract 1 from both sides: $$ 2x = -1 $$ Then divide both sides by 2: $$ x = -\frac{1}{2} $$ For the second equation \((x - 7) = 0\), add 7 to both sides: $$ x = 7 $$

Step 3: Write the solution

The solutions to the given equation are: $$ x = -\frac{1}{2} \quad \text{and} \quad x = 7 $$ Both of these values make the equation \((2x+1)(x-7)=0\) true.

Key Concepts

Factoring

Factoring is a critical skill when it comes to solving quadratic equations. But what does factoring mean? It involves breaking down a more complex expression into simpler parts, or "factors," that when multiplied together give back the original expression. This process makes it easier to solve the equation, especially when you're dealing with products like \((2x+1)(x-7)=0\).
For a quadratic expression, the goal is to express it as a product of two binomials. For example, \((2x+1)(x-7)\) is already factored into two binomials: \((2x+1)\) and \((x-7)\). This setup allows us to easily apply the next step, which is finding where this product equals zero.
Understanding how to factor is essential because it transforms a complex equation into manageable parts. This skill is not only useful in algebra problems but also valuable in more advanced math applications. Remember, practice makes perfect when it comes to mastering factoring.

Zero Product Property

The Zero Product Property is a fundamental principle in algebra that states if the product of two factors is zero, then at least one of the factors must be zero. This concept is key in solving quadratic equations. Let's apply it to our equation: \((2x+1)(x-7)=0\).
Here's how it works:

• If \((2x+1) = 0\), then the solution is found by solving for \(x\).
• If \((x-7) = 0\), same thing, solve for \(x\).

Each factor is treated independently, and solving each gives us possible solutions to the equation.
This property provides a straightforward method for determining solutions because it breaks down compound expressions into individual components that are easier to manage. Whenever you see a product equals zero, you can confidently split it into separate equations and solve each separately.This approach simplifies the entire solving process, reinforcing why the zero product property is so valuable when working through algebraic solutions.

Algebraic Solutions

Algebraic solutions help us find the values of variables that satisfy given equations. Solving equations like \((2x + 1)(x - 7) = 0\) requires a blend of strategies, including factoring and applying the zero product property. These steps lead us to algebraic solutions that tell us which values of \(x\) make the equation true.
For the equation at hand, our algebraic solution process includes:

• Setting each factor equal to zero.
• Solving these simpler equations to find potential values for \(x\).

This approach brings clarity to the process of solving more complex equations by reducing them into simpler pieces.
Using these algebraic methods, we calculated that \(x = -\frac{1}{2}\) and \(x = 7\). These solutions satisfy the original equation, ensuring that both \((2x+1)\) and \((x-7)\) each equal zero when their respective value for \(x\) is applied. Algebraic solutions are at the heart of solving equations in both basic and advanced mathematics. They guide us toward understanding and applying math in various contexts, reinforcing problem-solving skills that are essential in academic and real-world settings.