Problem 56 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use a graphing utility to approximate any solutions of the equation. [Remember to write the equation in the form \(f(x)=0.1\) $$|4 x+1|+2=8$$

Step-by-Step Solution

Answer

The solutions for the equation are \(x = 1.25\) and \(x = -1.75\).

Step 1: Convert absolute equation to standard form (i.e., \(f(x) = 0\))

The equation provided is \(|4x + 1| + 2 = 8\). To convert this equation into \(f(x) = 0\), start by subtracting 2 from both sides: \(|4x + 1| = 8 - 2\), simplifying this gives us \(|4x + 1| = 6\).

Step 2: Break down the absolute value equation

The absolute value equation \(|4x + 1| = 6\) could be broken down into two linear equations: \(4x + 1 = 6\) and \(4x + 1 = -6\).

Step 3: Solve the two linear equations

Solving the first linear equation (\(4x + 1 = 6\)): \ Start by subtracting 1 from both sides: \(4x = 6 - 1\), simplifying this gives us \(4x = 5\). Then, divide both sides by 4 to solve for \(x\), giving a solution \(x = 5/4\) or \(1.25\). \ Now, solve the second equation (\(4x + 1 = -6\)): Subtract 1 from both sides to get \(4x = -6 - 1\) or \(4x = -7\). Dividing both sides by 4 finds \(x = -7/4\) or \(-1.75\).

Step 4: Use a graphing utility

Draw the Graph for \(f(x) = |4x + 1| - 6\). The solutions achieved in step 3 will be presented in the graph as the x-intercepts where the graph crosses the x-axis. This verifies the solutions graphically.

Key Concepts

Graphing Utility Usage

Understanding how to use graphing utilities is crucial when dealing with absolute value equations. Let's take the given equation \( |4x + 1| + 2 = 8 \). First, it is important to convert it into the form \( f(x) = 0 \) to make it compatible with graphing tools. In this case, we subtract 2 from both sides to obtain \( f(x) = |4x + 1| - 6 \).

Once you have the correct form, input it into a graphing calculator or software. The graphing utility will provide a visual representation of the equation, showing where the function crosses the x-axis. These points of intersection are your solutions. Graphing utilities make identifying solutions more intuitive, especially for those who are visual learners. They also serve as a great tool to confirm the accuracy of your algebraic solutions.

Absolute Value Equation Conversion

An absolute value equation like \( |4x + 1| + 2 = 8 \) can seem daunting at first, but converting it into a solvable form is quite straightforward. The goal is to isolate the absolute value expression on one side to prepare for breaking it down further. By subtracting 2 from both sides of the given equation, we get \( |4x + 1| = 6 \).

This equation asserts that the expression within the absolute value bars is equal to 6 units from 0 on the number line, whether in the positive or negative direction. Hence, we proceed to split the equation into two separate linear equations to account for both possibilities, leading us towards finding that \( 4x + 1 \) can either equal 6 or -6.

Linear Equations Resolution

With the absolute value broken apart, we shift our focus to solving the resulting linear equations. Consider the two scenarios individually: \( 4x + 1 = 6 \) and \( 4x + 1 = -6 \).

For each equation, subtract 1 from both sides to isolate the term with \( x \). You'll end up with \( 4x = 5 \) for the first equation and \( 4x = -7 \) for the second. Dividing by 4 gives the solutions \( x = 1.25 \) and \( x = -1.75 \) respectively. These steps are crucial as they transform the problem from an absolute value equation into a pair of simple linear equations which can typically be solved with basic algebraic manipulation.

Graphical Verification of Solutions

Once algebraic solutions are found for an absolute value equation, it is wise to verify them graphically. A graphing utility can be used to visually represent the equation \( f(x) = |4x + 1| - 6 \) as a final check. On the graph, the x-intercepts—where the function crosses the x-axis—correspond to the solutions of the equation.

For the given problem, you should see the graph intersect the x-axis at \( x = 1.25 \) and \( x = -1.75 \) which align with the algebraic solutions. This visual confirmation is not only satisfying but also reinforces your understanding of the correlation between algebraic manipulation and its graphic representation.