Problem 30 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$x(12 x+11)=-2$$

Step-by-Step Solution

Answer

The real solutions are \(x = -\frac{1}{4}\) and \(x = -\frac{2}{3}\).

Step 1 - Expand the Equation

Start with the equation \(x(12x + 11) = -2\). Distribute \(x\) into the parenthesis to get the expanded form:\[12x^2 + 11x = -2.\]

Step 2 - Rearrange the Equation to Standard Form

Move \(-2\) to the left side of the equation to obtain the standard quadratic form:\[12x^2 + 11x + 2 = 0.\]

Step 3 - Identify a, b, and c

From the quadratic equation \(12x^2 + 11x + 2 = 0\), identify the coefficients: \(a = 12\), \(b = 11\), and \(c = 2\).

Step 4 - Use the Quadratic Formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots. Substitute \(a = 12\), \(b = 11\), and \(c = 2\) into the formula:\[x = \frac{-11 \pm \sqrt{11^2 - 4 \times 12 \times 2}}{2 \times 12}.\]

Step 5 - Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\) which is \(121 - 96 = 25\). Since the discriminant is positive, the equation has two real solutions.

Step 6 - Calculate the Solutions

Substitute the discriminant into the quadratic formula:\[x = \frac{-11 \pm \sqrt{25}}{24}.\]Solve for both values:\[x = \frac{-11 + 5}{24} = \frac{-6}{24} = -\frac{1}{4},\]\[x = \frac{-11 - 5}{24} = \frac{-16}{24} = -\frac{2}{3}.\]

Step 7 - Graph the Equation

Graph the quadratic equation \(12x^2 + 11x + 2 = 0\) to confirm the solutions. The graph will intercept the x-axis at \(x = -\frac{1}{4}\) and \(x = -\frac{2}{3}\). The graphical solution confirms the algebraic results.

Key Concepts

Graphical Solution

Graphical solutions allow us to visualize the roots of a quadratic equation. By graphing the equation, we can see where it crosses the x-axis. These points of intersection are the solutions to the equation.

• In our example, the equation is turned into its standard form: \(12x^2 + 11x + 2 = 0\).
• By plotting this equation as a parabola on a graph, we look for the points where the curve meets the x-axis.
• These intersections represent the real solutions of the equation.

In this case, upon graphing, we find the parabola intersects the x-axis at \(x = -\frac{1}{4}\) and \(x = -\frac{2}{3}\). This visual confirmation is crucial because it reinforces the validity of the solutions obtained algebraically.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Using the quadratic formula helps us find the roots of any quadratic equation, regardless of whether they can be easily factored. Let's see how it works:

• In our exercise, after rearranging, we have \(a = 12\), \(b = 11\), and \(c = 2\).
• By plugging these values into the formula, we start finding the roots using algebraic steps.

This formula is invaluable because it efficiently handles complex expressions and provides a direct route to solutions. Both real and imaginary solutions can arise from its use.

Discriminant

The discriminant is part of the quadratic formula under the square root sign: \(b^2 - 4ac\). It provides valuable information about the nature of the roots of the equation. Here's how it works:

• A positive discriminant indicates two distinct real roots.
• A discriminant of zero means there is exactly one real root (a repeated root).
• A negative discriminant reveals that there are no real roots but two complex roots.

In our example, the discriminant is calculated as \(11^2 - 4 \times 12 \times 2 = 25\). Since 25 is positive, it indicates that our quadratic equation has two distinct real roots, which are further supported by the graphical solution.

Real Solutions

Real solutions are the values of \(x\) for which the quadratic equation equals zero. They represent the points where the parabola crosses the x-axis. In this problem:

• The identified solutions are \(x = -\frac{1}{4}\) and \(x = -\frac{2}{3}\).
• These solutions are confirmed both algebraically, through the quadratic formula, and graphically, by plotting the equation.

Real solutions are vital as they offer a tangible connection between the equation and its graph. When a quadratic equation has real solutions, it means there are actual x-values that satisfy the equation of the parabola. These solutions are what we find and use while solving problems in algebra and calculus.