Problem 51 - Algebra for College Students

Question

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$(3 x-1)^{2}-16=0$$

Step-by-Step Solution

Answer

x = \frac{5}{3} or x = -1.

Step 1: Identify Equation Structure

The given equation is \((3x - 1)^2 - 16 = 0\). This can be viewed as a difference of squares. A difference of squares has the form \(a^2 - b^2 = 0\), which factors as \((a - b)(a + b) = 0\). Identify \(a\) and \(b\) to factor this equation.

Step 2: Rewrite as a Difference of Squares

Recognize that the equation \((3x - 1)^2 - 16 = 0\) is already a difference of squares where \(a = 3x - 1\) and \(b = 4\). The equation can be rewritten as \((3x - 1)^2 - 4^2 = 0\).

Step 3: Factor the Equation

Using the difference of squares formula \((a^2 - b^2 = (a - b)(a + b))\), factor the equation as:\((3x - 1 - 4)(3x - 1 + 4) = 0\)This simplifies to:\((3x - 5)(3x + 3) = 0\).

Step 4: Solve Each Factor for x

Set each factor equal to zero and solve for \(x\):1. For \(3x - 5 = 0\): - Add 5 to both sides: \(3x = 5\) - Divide by 3: \(x = \frac{5}{3}\)2. For \(3x + 3 = 0\): - Subtract 3 from both sides: \(3x = -3\) - Divide by 3: \(x = -1\).

Key Concepts

Difference of Squares

The difference of squares is a powerful technique often used in algebra to simplify certain types of quadratic expressions. It revolves around the principle that any expression in the form \(a^2 - b^2\) can be rewritten as \((a - b)(a + b)\). This technique is handy because it breaks down the expression into two simpler binomials.
For example, in the exercise you have the expression \((3x - 1)^2 - 16\). Notice that \(a\) is \((3x - 1)\) and \(b\) is \(4\), giving us the setup needed:

• \(a^2\) corresponds to \((3x - 1)^2\)
• \(b^2\) is \(16\) (since \(b = 4\))

When you factor this using the difference of squares formula, it simplifies the solving process considerably.

Solving Equations

Solving equations involves finding the values of the variable that make the equation true. In our exercise, you need to manage this through factoring techniques first. Once the equation is factored, as is the case with our quadratic equation, it's split into two linear equations.
After applying the difference of squares, we have the factors \((3x - 5)\) and \((3x + 3)\). You set each factor equal to zero because a product is zero if and only if at least one of the factors is zero.

• \(3x - 5 = 0\)
• \(3x + 3 = 0\)

By solving these simpler linear equations, we isolate \(x\) to find:

• From \(3x - 5 = 0\), solving for \(x\) gives \(x = \frac{5}{3}\).
• From \(3x + 3 = 0\), solving for \(x\) gives \(x = -1\).

These solutions satisfy the initial quadratic equation.

Factoring Quadratics

Factoring quadratics is a method used primarily for solving quadratic equations by rewriting them as products of binomials. It is an essential technique in algebra because it converts complex quadratic equations into more manageable ones.
In our example, the given expression \((3x - 1)^2 - 16 = 0\) was initially a quadratic in disguise. Recognizing it as a difference of squares allowed it to transform into two linear equations, which are much simpler to handle.
Here are key steps in factoring quadratics:

• Identify patterns such as a difference of squares or trinomials.
• Use relevant formulas to factorize effectively. In this case, it was \((a^2 - b^2)\).
• Split into simpler factors, as shown by \((3x - 1 - 4)(3x - 1 + 4) = 0\), leading to \((3x - 5)(3x + 3) = 0\).

This approach streamlines solving quadratics, reducing potential errors and making the calculation more straightforward for various equations.