Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$\frac{1}{6} x^{2}+\frac{1}{6} x=5$$

Step-by-Step Solution

Answer

The solutions are \( x = 5 \) and \( x = -6 \).

Step 1: Move All Terms to One Side of the Equation

First, let's rearrange the given equation to have one side equal to zero. The original equation is \( \frac{1}{6} x^{2} + \frac{1}{6} x = 5 \). Subtract 5 from both sides to get: \( \frac{1}{6} x^{2} + \frac{1}{6} x - 5 = 0 \).

Step 2: Multiply Through by 6 to Clear Fractions

To eliminate the fractions, multiply every term by 6, the denominator of the fractions. This gives us: \( x^2 + x - 30 = 0 \). We now have a quadratic equation in standard form.

Step 3: Use the Quadratic Formula to Find Solutions

We apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 1 \), and \( c = -30 \). First, calculate the discriminant: \( b^2 - 4ac = 1^2 - 4(1)(-30) = 1 + 120 = 121 \). Now, solve for \( x \): \( x = \frac{-1 \pm \sqrt{121}}{2 \times 1} = \frac{-1 \pm 11}{2} \).

Step 4: Calculate the Two Possible Solutions

Continuing from the quadratic formula, calculate each potential solution: \( x = \frac{-1 + 11}{2} = 5 \) and \( x = \frac{-1 - 11}{2} = -6 \). Thus, the solutions are \( x = 5 \) and \( x = -6 \).

Step 5: Verify Solutions Graphically

To support the solutions graphically, consider \( f(x) = \frac{1}{6} x^{2} + \frac{1}{6} x - 5 \). Plot this function and observe where it intersects the x-axis. The intersections at \( x = 5 \) and \( x = -6 \) confirm the solutions.

Key Concepts

Graphical Solutions

Graphing quadratic equations is a powerful way to visualize solutions to mathematical problems. A quadratic equation creates a parabola when plotted on a graph. The solutions to the equation are the x-values where the parabola intersects the x-axis. These points are known as the roots or solutions of the equation.

For the equation \( \frac{1}{6} x^2 + \frac{1}{6} x - 5 = 0 \), graphing involves plotting the corresponding function \( f(x) = \frac{1}{6} x^2 + \frac{1}{6} x - 5 \). By observing the graph:

• If the parabola touches the x-axis at two points, the equation has two distinct real solutions.
• If it only touches once, it has one real solution, known as a repeated root.
• If it does not touch the x-axis at all, the solutions are non-real (complex).

In this case, the graph intersects the x-axis at \( x = 5 \) and \( x = -6 \), confirming these are the solutions to the equation.

Quadratic Formula

A fundamental method for solving quadratic equations is the quadratic formula. This formula provides a direct way to find the solutions of any quadratic equation in the form \( ax^2 + bx + c = 0 \). The formula is expressed as:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here's how you can use the formula:

• Identify the coefficients \( a \), \( b \), and \( c \) from the equation.
• Compute the discriminant \( b^2 - 4ac \) to determine the nature of the roots.
• Substitute the values into the formula to calculate the possible values of \( x \).

For the equation \( x^2 + x - 30 = 0 \), the values are \( a = 1 \), \( b = 1 \), and \( c = -30 \). Plugging these into the quadratic formula gives \( x = 5 \) and \( x = -6 \), the valid solutions.

Discriminant Calculation

The discriminant is a part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is the expression located under the square root in the formula:

\( b^2 - 4ac \)

The discriminant offers insight into whether the solutions are real or complex:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is exactly one real root, indicating a perfect square trinomial.
• If \( b^2 - 4ac < 0 \), the roots are complex and not real numbers.

For \( x^2 + x - 30 = 0 \), the calculation \( 1^2 - 4(1)(-30) = 121 \) shows that the discriminant is positive, indicating two distinct real solutions.

Standard Form of Quadratic Equation

The standard form of a quadratic equation is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Converting an equation to this form is essential for utilizing the quadratic formula or other methods of solution.

In this problem, the original equation \(\frac{1}{6} x^2 + \frac{1}{6} x = 5 \) was converted to standard form by rearranging it as \( x^2 + x - 30 = 0 \) after multiplying through by 6 to clear fractions.

Simplifying the equation to standard form involves:

• Moving identical terms to one side so that the expression equals zero.
• Clearing any coefficients from the variable terms, typically by multiplying through by a denominator if fractions are involved.

Having the equation in standard form allows easy identification and application of algebraic tools to solve the problem efficiently.