Problem 37 - A Graphical Approach to Precalculus with Limits

Question

Use an analytic method to solve each equation in part (a). Support the solution with a graph. Then use the graph to solve the inequalities in parts (b) and (c). (a) \(\sqrt{5 x+1}+2=2 x\) (b) \(\sqrt{5 x+1}+2>2 x\) (c) \(\sqrt{5 x+1}+2<2 x\)

Step-by-Step Solution

Answer

(a) Solution is x = 3; (b) \([0, 3)\); (c) \((3, \infty)\).

Step 1: Rearrange the Equation

Start with the equation \(\sqrt{5x+1} + 2 = 2x\). Subtract 2 from both sides to isolate the square root: \(\sqrt{5x+1} = 2x - 2\).

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root: \((\sqrt{5x+1})^2 = (2x - 2)^2\). This simplifies to \(5x + 1 = 4x^2 - 8x + 4\).

Step 3: Simplify and Rearrange

Bring all terms to one side of the equation to form a quadratic equation: \(4x^2 - 13x + 3 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -13\), and \(c = 3\). Calculate the discriminant:\(b^2 - 4ac = 169 - 48 = 121\). So, \(x = \frac{13 \pm 11}{8}\). The solutions are \(x = 3\) and \(x = \frac{1}{4}\).

Step 5: Verify the Solutions

Substitute \(x = 3\) into the original equation: \(\sqrt{5(3)+1} + 2 = 2(3)\) simplifies to \(\sqrt{16} + 2 = 6\), true as \(4 + 2 = 6\). Substitute \(x = \frac{1}{4}\) into the original equation: \(\sqrt{5(\frac{1}{4}) + 1} + 2 = 2(\frac{1}{4})\), which simplifies to \(\sqrt{\frac{5}{4} + 1} + 2 = \frac{1}{2}\), not true as \(\sqrt{2.25} + 2 eq 0.5\). Only \(x = 3\) is valid.

Step 6: Graph the Functions

Plot the functions \(y_1 = \sqrt{5x+1} + 2\) and \(y_2 = 2x\) on a graph to visually verify the solutions. They intersect at \((3,6)\).

Step 7: Solve the Inequality \(\sqrt{5x+1}+2>2x\)

From the graph, \(y_1 = \sqrt{5x+1} + 2\) is above \(y_2 = 2x\) when \(0 \leq x < 3\). Thus, the solution to the inequality is \([0, 3)\).

Step 8: Solve the Inequality \(\sqrt{5x+1}+2<2x\)

From the graph, \(y_1 = \sqrt{5x+1} + 2\) is below \(y_2 = 2x\) when \(x > 3\). Thus, the solution to the inequality is \((3, \infty)\).

Key Concepts

Quadratic Equation

A quadratic equation is any equation that can be rearranged into the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are known values where \(a eq 0\) because if \(a\) were zero, the equation would be linear, not quadratic.
Quadratic equations are commonplace in algebra and can represent various relationships in the real world, such as the path of a thrown ball or optimizing areas. In this exercise, the process begins with transforming the original equation \(\sqrt{5x+1} + 2 = 2x\) into a quadratic form. We do this by squaring both sides, which removes the square root and gives us \(5x + 1 = 4x^2 - 8x + 4\).
This equation is further rearranged to \(4x^2 - 13x + 3 = 0\), a standard quadratic form. Solving this equation involves applying the quadratic formula:

• Identify \(a = 4\), \(b = -13\), and \(c = 3\).
• Calculate the discriminant \(b^2 - 4ac = 121\).
• Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find solutions \(x = 3\) and \(x = \frac{1}{4}\).

This process highlights the versatility of the quadratic formula in finding solutions to polynomial equations.

Graphing Functions

Graphing functions involves drawing their respective curves on a coordinate plane to visualize solutions and relationships between variables. This is a powerful tool in algebra, providing an intuitive understanding far beyond mere calculations.
In the solved exercise, two functions are graphed: \(y_1 = \sqrt{5x+1} + 2\) and \(y_2 = 2x\). Plotting these helps verify the intersections and inequality solutions. The graphs intersect at \((3,6)\), confirming that \(x = 3\) is a valid solution for \(\sqrt{5x+1} + 2 = 2x\).
When graphing:

• Identify key points, such as intersections, zeros, and asymptotes, if any.
• Ensure clarity in labeling axes and scaling for accurate representation and interpretation.
• Use intersection points to solve equations and inequalities effectively.

Through these steps, students can connect algebraic solutions to visual representations, enhancing comprehension and recall of concepts related to function behavior.

Inequalities

Inequalities, unlike equations, describe a range of values rather than a specific solution. Understanding inequalities is crucial, especially when their graphical analyses help determine the intervals where a function's value lies above or below another.
In this exercise, we determine where \(y_1 = \sqrt{5x+1} + 2\) is greater than \(y_2 = 2x\). Using the graph, it's clear that \(y_1 > y_2\) for \(0 \leq x < 3\). This yields the inequality solution \([0, 3)\).
Similarly, for \(y_1 < y_2\), observe that \(x > 3\) satisfies this condition, which translates to the solution \((3, \infty)\). Key points when solving inequalities include:

• Recognize that graphical intersections serve as boundaries for inequalities.
• Read the graph accurately to find where one function dominates the other.
• Express solutions in interval notation to clearly define the range of valid \(x\) values.

Mastering inequalities ensures that students can handle a variety of mathematical and real-world scenarios with precision and confidence.