Problem 19 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ 6 x(2 x-5)=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = \frac{5}{2}\).

Step 1: Identifying the Zero Product Property

The equation given is in the factored form: \(6x(2x-5)=0\). According to the zero product property, if a product of two factors equals zero, at least one of the factors must be zero.

Step 2: Apply Zero Product Property

We set each factor to zero separately: 1. \(6x = 0\) 2. \(2x - 5 = 0\). We will solve each equation to find the values of \(x\) that make the equation true.

Step 3: Solve First Equation

Solve the equation \(6x = 0\) by dividing both sides by 6: \[ x = \frac{0}{6} = 0 \]. Thus, one solution is \(x = 0\).

Step 4: Solve Second Equation

Solve the equation \(2x - 5 = 0\) by isolating \(x\). Add 5 to both sides: \[ 2x = 5 \]. Divide both sides by 2: \[ x = \frac{5}{2} \]. Thus, another solution is \(x = \frac{5}{2}\).

Step 5: Combine Solutions

The solutions to the equation \(6x(2x-5)=0\) are \(x = 0\) and \(x = \frac{5}{2}\).

Key Concepts

Factored Form

The term **factored form** refers to the way we express an equation as a product of its factors, especially in polynomial equations like quadratic ones. For instance, consider the equation \(6x(2x-5) = 0\). This equation is already in factored form, which means it's written as a multiplication of different expressions such that if you open them up using distributive property, you get back the original polynomial.
Why is this form so useful, you might wonder? Because, according to the **zero product property**, if the product of factors is zero, then at least one of those factors must be zero. This allows us to easily identify the potential solutions to the equation by setting each factor equal to zero.
In our example, the factors are:\(6x\) and \(2x-5\). You set each equal to zero separately to solve for the variable \(x\). Factoring an equation makes it much simpler to solve because it breaks the equation down into manageable parts.

Solving Quadratic Equations

Quadratic equations are equations that can be expressed in the form \(ax^2 + bx + c = 0\). They can often be solved more easily when converted into factored form. Let's revisit our example equation \(6x(2x-5) = 0\).
When solving quadratic equations, you want to find the values of \(x\) that satisfy the equation. We do this by applying the zero product property. First, identify the factors:

• \(6x = 0\)
• \(2x - 5 = 0\)

The next step is simply solving these individual equations:
For \(6x = 0\), divide both sides by 6, resulting in \(x = 0\).
For \(2x - 5 = 0\), add 5 to both sides to get \(2x = 5\), then divide by 2 to solve \(x = \frac{5}{2}\).
Thus, by transitioning from the expanded form of a quadratic equation to its factored form, you can easily solve for \(x\) using straightforward algebraic manipulation.

Algebraic Solutions

When dealing with **algebraic solutions**, we aim to isolate the variable (often \(x\)) to find its values. Starting from the factored form is a critical step as it lays the groundwork for these solutions through simple algebraic operations.
In our example, transitioning to algebraic solutions involved solving two simpler equations:

• First, \(6x = 0\) provided the direct solution of \(x = 0\).
• Second, \(2x - 5 = 0\) required basic algebra like addition and division to find \(x = \frac{5}{2}\).

These operations demonstrate how algebraic techniques can turn a seemingly complex quadratic equation into a series of manageable tasks, each providing parts of the complete solution.
A key takeaway is understanding the balance and systematic approach required. Start by identifying the factors, apply algebraic techniques to each, and gather the results. This method simplifies the process, ensuring you have comprehensive solutions to quadratic problems.