Problem 68 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the discriminant to determine the number of real solutions of the equation. $$ 25 x^{2}-80 x+64=0 $$

Step-by-Step Solution

Answer

For the given quadratic equation \(25x^2 - 80x + 64 = 0\), we find the discriminant using the formula \(\Delta = b^2 - 4ac = (-80)^2 - 4(25)(64) = 6400 - 6400 = 0\). Since the discriminant is 0, the equation has exactly one real solution.

Step 1: Identify the values of a, b, and c

We are given the equation \(25x^2 - 80x + 64 = 0\). This equation is in the form of a quadratic equation, \(ax^2 + bx + c = 0\). Therefore, we can identify the values of a, b, and c as follows: - a = 25 - b = -80 - c = 64

Step 2: Calculate the discriminant

To calculate the discriminant, we'll use the formula \(\Delta = b^2 - 4ac\). Plug the values of a, b, and c into the formula: \(\Delta = (-80)^2 - 4(25)(64)\)

Step 3: Simplify the expression

Simplify the expression for the discriminant: \(\Delta = 6400 - 4(25)(64) = 6400 - 6400\)

Step 4: Evaluate the discriminant

Evaluate the discriminant: \(\Delta = 0\)

Step 5: Determine the number of real solutions

Since the discriminant is equal to 0 (\(\Delta = 0\)), the quadratic equation \(25x^2 - 80x + 64 = 0\) has exactly one real solution.

Key Concepts

Discriminant

The discriminant is a powerful component found in the context of quadratic equations. It gives us valuable insight into the nature of the solutions without actually solving the equation.

The discriminant is denoted as \(\Delta\) and is calculated from the coefficients of a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) represent real numbers, with \(a \eq 0\). The discriminant formula is \(\Delta = b^2 - 4ac\).

The value of the discriminant reveals the number and type of solutions:

• If \(\Delta > 0\), the equation has two distinct real solutions.
• If \(\Delta = 0\), there is exactly one real solution.
• If \(\Delta < 0\), there are no real solutions; instead, there are two complex solutions.

In our example, the discriminant \(\Delta = (-80)^2 - 4(25)(64)\) simplifies to \(\Delta = 6400 - 6400 = 0\), indicating that the quadratic equation has a single unique real solution.

Quadratic Formula

To solve a quadratic equation, which is an equation of the second degree, we often use the quadratic formula. The general equation \(ax^2 + bx + c = 0\) can be solved for \(x\) using this straightforward formula: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]This formula elucidates that the solutions for \(x\) depend on the values of the coefficients \(a\), \(b\), and \(c\). Additionally, the \(\pm\) symbol represents the two possible values for \(x\), stemming from the square root of the discriminant. If the discriminant is negative, the square root becomes imaginary, leading to complex solutions.

Using the quadratic formula simplifies the process of finding solutions and ensures that no solution is overlooked, making it an essential tool in algebra.

Real Number Solutions

Quadratic equations, depending on their discriminant, may have real number solutions or complex solutions. Real number solutions are the possible values for \(x\) that satisfy the equation when graphed on a standard x-y coordinate plane; they are points where the parabola (the graph of a quadratic function) intersects the x-axis.

For our quadratic equation \(25x^2 - 80x + 64 = 0\), the discriminant equals to zero (\(\Delta = 0\)). This reveals that the parabola touches the x-axis at exactly one point. Hence, it has a single real solution, which could also be interpreted as the vertex of the parabola lying on the x-axis. In this particular case, the quadratic formula will yield one value for \(x\) that will be a 'double root', meaning that both solutions of the equation are identical.

This concept ensures that even when intricate algebra is involved, by determining the nature and number of solutions, one can have a clearer perspective on the structure of the quadratic equation before attempting to solve it.