Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Use the discriminant to determine the number of real solutions of the quadratic equation. \(\frac{1}{5} x^{2}+\frac{6}{5} x-8=0\)

Step-by-Step Solution

Answer

The quadratic equation \(\frac{1}{5} x^{2}+\frac{6}{5} x-8=0\) has two distinct real solutions.

Step 1: Identifying a, b and c

In the given equation \(\frac{1}{5} x^{2}+\frac{6}{5} x-8=0\), the values of a, b and c (the coefficients of \(x^2\), x, and the constant respectively) can be identified as: \(a = \frac{1}{5}\), \(b = \frac{6}{5}\), and \(c = -8\).

Step 2: Calculating the Discriminant

The discriminant D is calculated by using the formula \(D = b^2 - 4ac\). Substituting the given values in, we get \(D = \left(\frac{6}{5}\right)^2 - 4 * \frac{1}{5} * -8\). We can simplify this to \(D = \frac{36}{25} + \frac{32}{5} = \frac{196}{25}\).

Step 3: Evaluating the Discriminant

The discriminant D is \(\frac{196}{25}\), which is greater than 0. Therefore, the given quadratic equation has two distinct real roots.

Key Concepts

Quadratic Equation Solutions

When it comes to solving quadratic equations, which are of the form \( ax^2 + bx + c = 0 \), the solutions hinge on the nature of the discriminant. This part of the quadratic formula, represented by \( D \text{ or } \Delta \text{ (Delta)} \text{, determines whether you're going to have two real and distinct solutions, one real and repeated solution, or two complex solutions.} \)

The magical formula to find the roots (solutions) is \( x = \frac{-b \pm \sqrt{D}}{2a} \text{, with the }\pm \text{ indicating that you might have two solutions.} \) For our exercise \( \frac{1}{5} x^{2}+\frac{6}{5} x-8=0 \), we can apply this formula once we've calculated the discriminant, D. As the exercise revealed, since D is greater than zero, we expect two distinct real solutions, which we can find by plugging the values into this formula.

Calculating Discriminant

The discriminant of a quadratic equation offers insight into the nature of its roots without requiring you to actually solve the equation. The formula for the discriminant is \( D = b^2 - 4ac \). A positive discriminant indicates two real and distinct solutions, a discriminant of zero suggests one real and repeated solution, and a negative discriminant points to complex solutions with no real part.

For example, in our exercise, we determined that \( a = \frac{1}{5} \), \( b = \frac{6}{5} \), and \( c = -8 \). Upon calculation, the discriminant turned out to be \( \frac{196}{25} \), which is positive. This tells us, even before any actual solving, that we're going to find real and separate solutions for the quadratic equation. Such preemptive knowledge is powerful in mathematical problem solving as it sets expectations and guides subsequent steps.

Real Roots of Quadratic

The roots of a quadratic equation are the values of \( x \) that make the equation true, and they are derived from where the graph of the quadratic equation crosses the x-axis. A positive discriminant, as seen with \( \frac{196}{25} \), confirms that we have two distinct points where our parabola intersects the x-axis, hence two real roots.

These roots can represent various scenarios in real-world problems, from projections in sales to the calculation of trajectories. Understanding the significance of the discriminant helps us predict the number and nature of these roots. When a discriminant is positive and a perfect square, as in our example, the roots are not only real and distinct but also rational numbers. This can greatly simplify further applications of these roots in practical examples, allowing for precise and meaningful interpretations.