Problem 17 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$40 x-16 x^{2}=0$$

Step-by-Step Solution

Answer

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

Step 1: Write the Equation in Standard Form

The equation is given as \(40x - 16x^2 = 0\). Let's rearrange it to be in the form \(ax^2 + bx + c = 0\). So, it becomes \(-16x^2 + 40x = 0\).

Step 2: Factor Out the Greatest Common Factor (GCF)

Identify and factor out the greatest common factor from each term. Here, the GCF of \(-16x^2 + 40x\) is \(8x\). Thus, the equation becomes \(8x(-2x + 5) = 0\).

Step 3: Apply the Zero Product Property

The Zero Product Property states that if \(ab = 0\), then \(a = 0\) or \(b = 0\). Apply it here: Either \(8x = 0\) or \(-2x + 5 = 0\).

Step 4: Solve Each Equation Separately

Firstly solve \(8x = 0\):1. Divide both sides by 8: \[x = 0\]Next, solve \(-2x + 5 = 0\):1. Subtract 5 from both sides: \[-2x = -5\]2. Divide by \(-2\): \[x = \frac{5}{2}\]

Step 5: Conclude the Solutions

The solutions to the quadratic equation \(-16x^2 + 40x = 0\) are \(x = 0\) and \(x = \frac{5}{2}\).

Key Concepts

Zero Product Property

The Zero Product Property is a helpful principle when dealing with quadratic equations like the one in our exercise. It's a straightforward concept: if the product of two numbers (say \(a\) and \(b\)) equals zero, then either \(a = 0\) or \(b = 0\) or both. This property is what makes factoring such a powerful tool for solving quadratic equations.

In general, when you have a factored expression set equal to zero, you can write two separate equations. Each factor is set to zero individually, such as \(ab = 0\) resulting in \(a = 0\) and \(b = 0\).

This is exactly what we did in the exercise. After factoring the equation to get \(8x(-2x + 5) = 0\), we applied the Zero Product Property:

• For the first factor: \(8x = 0\)
• For the second factor: \(-2x + 5 = 0\)

This process allows us to easily solve for the variable \(x\).

Standard Form of Quadratic Equations

Quadratic equations typically appear in what's called 'standard form'. The standard form is \(ax^2 + bx + c = 0\). Recognizing this is crucial because it sets the stage for factoring or applying other solving techniques.

In our specific exercise, the equation was initially given as \(40x - 16x^2 = 0\). Notice it wasn’t in the neat \(ax^2 + bx + c\) format, so we had to rearrange it to \(-16x^2 + 40x = 0\).

Getting the equation in standard form allows us to clearly see all terms and their coefficients:

• \(a = -16\)
• \(b = 40\)
• \(c = 0\)

With this setup, we can now move forward to factoring the equation effectively.

Greatest Common Factor (GCF)

Finding the Greatest Common Factor (GCF) is a key step in simplifying expressions. It’s the largest number that divides exactly into each term of the expression, and it’s an essential part of factoring a quadratic equation.

For our exercise, the quadratic equation \(-16x^2 + 40x = 0\) had terms that shared a common factor. By identifying that the GCF was \(8x\), we were able to factor the equation into \(8x(-2x + 5) = 0\).

This step simplified the process enormously. Here's how you typically find the GCF:

• List the factors of each coefficient and the variables separately.
• Identify the largest factor common to all terms involved.

Pulling out the GCF helps break down the equation into a simpler, factored form, ready for the application of the zero product property.

Quadratic Equation Solutions

The ultimate goal in solving a quadratic equation is to find the values of \(x\) that satisfy the equation. In our exercise, after factoring and applying the zero product property, we formed two separate simple equations.

The solutions, \(x = 0\) and \(x = \frac{5}{2}\), were derived from these simpler equations:

• For \(8x = 0\), dividing by 8 gives us \(x = 0\).
• For \(-2x + 5 = 0\), rearranging and dividing gives us \(x = \frac{5}{2}\).

These solutions mean that plugging either value for \(x\) back into the original quadratic equation will satisfy the equation, making it true. This illustrates how effective factoring, along with the zero product property, can be in finding solutions to quadratic equations.