Problem 15 - Applied Calculus

Question

Solve each equation by factoring. [Hint for: First factor out a fractional power.] $$ 2 x^{3}+18 x=12 x^{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = 3\).

Step 1: Move All Terms to One Side

Start by rewriting the given equation so that all terms are on one side. Subtract \(12x^2\) from both sides to get: \[2x^3 + 18x - 12x^2 = 0.\]

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

Identify the greatest common factor of the terms \(2x^3 + 18x - 12x^2 \). The GCF is \(2x\). Factor \(2x\) out from each term: \[2x(x^2 - 6x + 9) = 0.\]

Step 3: Factor the Quadratic Expression

Factor the quadratic expression \(x^2 - 6x + 9\). Recognize this as a perfect square trinomial: \(x^2 - 6x + 9 = (x-3)^2\). So, the equation becomes: \[2x(x-3)^2 = 0.\]

Step 4: Set Each Factor to Zero

Solve for \(x\) by setting each factor equal to zero: \(2x = 0\) and \((x-3)^2 = 0\).

Step 5: Solve the Linear and Quadratic Equations

Solve the linear equation \(2x = 0\) to find \(x = 0\). Solve the equation \((x-3)^2 = 0\) by taking the square root of both sides: \(x - 3 = 0\), giving \(x = 3\).

Step 6: Combine Solutions

Combine the solutions from the linear and quadratic equations. The solutions are \(x = 0\) and \(x = 3\).

Key Concepts

Understanding Quadratic Equations

Quadratic equations are polynomials of degree 2, generally represented as \( ax^2 + bx + c = 0 \). They are essential in mathematics because they represent the simplest non-linear equation. A quadratic equation forms a parabola when graphed on the Cartesian coordinate system. The solutions or roots of the quadratic equation correspond to the points where the parabola intersects the x-axis. Understanding how to solve these equations is crucial for a wide array of mathematical applications and problem-solving scenarios.

Quadratic equations can be solved by:

• Factoring
• Using the quadratic formula
• Completing the square

Factoring is a preferred method when the equation can be easily decomposed into binomial products, as shown in the problem you've started with.

Perfect Square Trinomials Simplified

A perfect square trinomial is a special form of quadratic expression which is the square of a binomial. It takes the pattern \( (a+b)^2 = a^2 + 2ab + b^2 \). Recognizing a perfect square trinomial can simplify solving quadratic equations.

For the trinomial \( x^2 - 6x + 9 \), notice that it follows the pattern of \( (x-3)^2 \) because:

• \( a^2 = x^2 \),\( a = x \)
• \( b^2 = 9 \), \( b = 3 \)
• \( 2ab = 6x \)

Therefore, \( x^2 - 6x + 9 \) can be rewritten as \( (x-3)^2 \), greatly simplifying the process of solving the equation as you only need to solve \( (x-3)^2 = 0 \) instead of dealing with a more complex expression.

The Greatest Common Factor (GCF)

Finding the greatest common factor (GCF) is a pivotal step in simplifying equations through factoring. The GCF is the highest number or expression that can exactly divide all terms in the polynomial. By identifying and factoring out the GCF, we simplify the equation, making further factoring more manageable.

In the equation \( 2x^3 + 18x = 12x^2 \), after rearranging, you get \( 2x^3 + 18x - 12x^2 = 0 \). Here, the GCF is \( 2x \), which helps to simplify the equation to \( 2x(x^2 - 6x + 9) = 0 \). This step sets the stage for any further factoring or solving, by helping in reducing complexity right at the beginning.

Solving Linear Equation Solutions

When solving quadratic equations by factoring, you often encounter linear equation solutions. These are equations of the form \( ax + b = 0 \), which are straightforward to solve. The solution represents the values of \( x \) where the equation equals zero.

In the original exercise, after factoring, we get \( 2x(x-3)^2 = 0 \). Solving \( 2x = 0 \) is a linear equation. To solve this, you simply divide both sides by 2, resulting in \( x = 0 \). This quick step is often overlooked, yet critical, as it provides one of the roots or solutions to the equation.

Linear equations often provide instant solutions when they appear in the process of factoring complex equations, offering a quick step towards the resolution of larger problems.