Problem 83 - Algebra A Combined Function

Question

Solve each equation. $$ (x-3)(3 x+4)=(x+2)(x-6) $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = \frac{1}{2}\).

Step 1: Expand Both Sides

First, let's expand both sides of the equation. Expand the left side: \((x-3)(3x+4) = x(3x+4) - 3(3x+4) = 3x^2 + 4x - 9x - 12 = 3x^2 - 5x - 12\).Expand the right side: \((x+2)(x-6) = x(x-6) + 2(x-6) = x^2 - 6x + 2x - 12 = x^2 - 4x - 12\). Now the equation is \(3x^2 - 5x - 12 = x^2 - 4x - 12\).

Step 2: Move All Terms to One Side

To simplify, move all terms to one side of the equation. Subtract \(x^2\), add \(4x\), and add \(12\) to both sides: \(3x^2 - 5x - 12 - x^2 + 4x + 12 = 0\).This simplifies to \(2x^2 - x = 0\).

Step 3: Factor the Quadratic Equation

Now, let's factor the quadratic expression.\(2x^2 - x = x(2x - 1) = 0\).This gives us two factors: \(x = 0\) or \(2x - 1 = 0\).

Step 4: Solve for x

For the equation \(x = 0\), we have one solution, \(x = 0\).For the equation \(2x - 1 = 0\), add 1 and divide by 2:\(2x = 1\) \(x = \frac{1}{2}\).Thus, the solutions are \(x = 0\) and \(x = \frac{1}{2}\).

Key Concepts

Factoring

Factoring is a fundamental skill in solving quadratic equations. It involves breaking down a complex expression into simpler components, called factors. In this exercise, after expanding and simplifying the equation to reach the expression \(2x^2 - x = 0\), we use factoring to simplify further.
To factor \(2x^2 - x\), we notice both terms share the common factor \(x\). This means we can "factor out" \(x\) from each term, resulting in \(x(2x - 1)\). Factors like this allow us to set each part equal to zero and find solutions easily.
Factoring requires keen observation to identify common terms or patterns like square terms or differences, making the equation manageable. Remember: always look for the greatest common factor or any notable patterns like perfect squares or difference of squares.

Solution Set

The solution set of an equation represents all possible values that satisfy the equation. In quadratic equations, we often find two solutions. After factoring \(2x^2 - x = 0\) into \(x(2x - 1) = 0\), we use the zero-product property. This property states if a product equals zero, then at least one of the factors must equal zero.
Let's examine each factor:

• \(x = 0\): A straightforward substitution reveals one solution.
• \(2x - 1 = 0\): Add 1 to both sides and divide by 2, resulting in \(x = \frac{1}{2}\).

Hence, the solution set is \(x = 0\) and \(x = \frac{1}{2}\). Remember, the solution set contains all the values for \(x\) that make the original equation true. Identifying these solutions involves breaking down and setting each factored component to zero.

Expansion of Expressions

Before solving quadratic equations, expanding expressions is often the first step. Expansion simplifies complex multiplications into standard polynomial forms.
In this exercise, we start with expressions like \((x-3)(3x+4)\) and \((x+2)(x-6)\). Expansion involves distributing each term in the first bracket across every term in the second bracket.
Here's how that works for our expressions:

• Left side: \((x-3)(3x+4)\) becomes \(3x^2 + 4x - 9x - 12 = 3x^2 - 5x - 12\).
• Right side: \((x+2)(x-6)\) expands to \(x^2 - 6x + 2x - 12 = x^2 - 4x - 12\).

By expanding both sides, we can easily see the polynomial form \(3x^2 - 5x - 12 = x^2 - 4x - 12\), which leads the way toward combining like terms and further simplification. This transformation is crucial for setting the stage standard quadratic equations can tackle.