Problem 58 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$-5 x^{2}+28 x+12=0$$

Step-by-Step Solution

Answer

The solutions are \(x = -0.4\) and \(x = 6\), shown graphically as the x-intercepts of the parabola.

Step 1: Identify the Type of Equation

The given equation is \[-5x^2 + 28x + 12 = 0\].This is a quadratic equation because it has the form \(ax^2 + bx + c = 0\), where \(a = -5\), \(b = 28\), and \(c = 12\).

Step 2: Use the Quadratic Formula

The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substitute the values of \(a\), \(b\), and \(c\) into the formula: \[x = \frac{-28 \pm \sqrt{28^2 - 4(-5)(12)}}{2(-5)}\].

Step 3: Calculate the Discriminant

The discriminant is the part under the square root: \(b^2 - 4ac\).Calculate it: \[28^2 - 4(-5)(12) = 784 + 240 = 1024\]. Since the discriminant is positive, there are two real solutions.

Step 4: Solve for the Roots

Continue using the quadratic formula: \[x = \frac{-28 \pm \sqrt{1024}}{-10}\]. Since \(\sqrt{1024} = 32\), we have:\[x = \frac{-28 \pm 32}{-10}\]. Calculate each solution:- For \(+32\) : \(x = \frac{-28 + 32}{-10} = \frac{4}{-10} = -0.4\).- For \(-32\) : \(x = \frac{-28 - 32}{-10} = \frac{-60}{-10} = 6\).

Step 5: Graphical Representation

Graph the equation \(y = -5x^2 + 28x + 12\). The graph is a downward-opening parabola with roots at \(x = -0.4\) and \(x = 6\). These points are where the graph intersects the x-axis.

Key Concepts

Quadratic Formula

Quadratic equations are pivotal in mathematics as they represent the second-degree polynomial equations in the form of \(ax^2 + bx + c = 0\). The quadratic formula provides a straightforward method to find the solutions of these equations. It's expressed as: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is handy because it can solve any quadratic equation. By plugging the coefficients \(a\), \(b\), and \(c\) into this formula, you can effortlessly find the solutions, also known as roots.

• \(a\) represents the coefficient of \(x^2\).
• \(b\) is the coefficient of \(x\).
• \(c\) stands for the constant term.

In our example, plug in \(a = -5\), \(b = 28\), and \(c = 12\) to find the solutions of the equation \(-5x^2 + 28x + 12 = 0\). The quadratic formula unveils the roots by substituting these parameters and simplifying.

Discriminant

A critical part of the quadratic formula is the discriminant, \(b^2 - 4ac\). It determines the nature and number of roots of a quadratic equation, offering insights even before any calculations.

• Positive Discriminant: If \(b^2 - 4ac > 0\), there are two distinct real roots.
• Zero Discriminant: If \(b^2 - 4ac = 0\), there is one real root (or two identical real roots).
• Negative Discriminant: If \(b^2 - 4ac < 0\), there are no real roots, only complex ones.

In our case, the equation \(-5x^2 + 28x + 12 = 0\) has a discriminant of 1024, which is positive. Therefore, there are two distinct real solutions for this quadratic equation. The discriminant gives a quick preview of what to expect from the equation: real or complex roots.

Graphing Parabolas

Graphing a quadratic equation visually illustrates the roots and the shape of the solution. A quadratic equation represents a parabola on a coordinate plane. The general form \(ax^2 + bx + c = 0\) determines the parabola's shape and direction.For our example, \(y = -5x^2 + 28x + 12\), the graph is a downward-opening parabola because the coefficient \(a = -5\) is negative.

• The parabola will intersect the x-axis at the equation's real solutions, known as the roots. Here, it intersects at \(x = -0.4\) and \(x = 6\).
• The vertex of the parabola, being the maximum point in the downward direction, can be calculated using \(x = -\frac{b}{2a}\).
• Understanding vertex positions and the direction helps in sketching and analyzing parabolas.

Graphing represents not only solutions but gives perspective about maximum and minimum values and the overall trends of the equation across its domain.