Problem 69 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the discriminant to determine the number of real solutions of the equation. $$ \frac{6}{k^{2}}+\frac{1}{k}-2=0 $$

Step-by-Step Solution

Answer

The given equation is \(\frac{6}{k^2} + \frac{1}{k} - 2 = 0\). First, rewrite the equation as a quadratic equation: \(-2k^2 + k + 6 = 0\). Next, calculate the discriminant: \(D = 49\). Since \(D > 0\), the equation has 2 distinct real solutions.

Step 1: Rewrite the given equation as a quadratic equation

To rewrite the given equation as a quadratic equation, we need to manipulate it to get an equation in the form of \(ax^2 + bx + c = 0\) where a, b, and c are constants: $$ \frac{6}{k^2} + \frac{1}{k} - 2 = 0 $$ $$ \frac{6 + k}{k^2} - 2 = 0 $$ We can multiply both sides by \(k^2\) to eliminate the denominators: $$ 6 + k - 2k^2 = 0 $$ Now, the equation is in quadratic form: $$ -2k^2 + k + 6 = 0 $$ The coefficients are a = -2, b = 1, and c = 6.

Step 2: Calculate the discriminant of the quadratic equation

The discriminant, D, is calculated using the formula: \(D = b^2 - 4ac\). We need to substitute the values of a, b, and c into the formula: $$ D = (1)^2 - 4(-2)(6) $$ $$ D = 1 + 48 $$ $$ D = 49 $$

Step 3: Determine the number of real solutions based on the value of the discriminant

The value of the discriminant determines the number of real solutions for the quadratic equation: - If \(D > 0\), the equation has 2 distinct real solutions. - If \(D = 0\), the equation has 1 real solution (a repeated root). - If \(D < 0\), the equation has no real solutions, only complex solutions. In our case, the discriminant \(D = 49\), which is greater than 0. Therefore, the equation \(\frac{6}{k^2} + \frac{1}{k} - 2 = 0\) has 2 distinct real solutions.

Key Concepts

Real Solutions of Quadratic Equations

Understanding the concept of real solutions in quadratic equations is crucial for students tackling algebra. When a quadratic equation is presented as an equation where the highest exponent of the variable is 2, it can take the form of \(ax^2 + bx + c = 0\). The number of real solutions such a quadratic equation has, can be determined by looking at the discriminant, which is a part of the quadratic formula.
To find the discriminant, we use the formula \(D = b^2 - 4ac\). The value of the discriminant will tell us how many real solutions the equation has:

• If \(D > 0\), there are two distinct real solutions.
• If \(D = 0\), there is exactly one real solution.
• If \(D < 0\), there are no real solutions, only complex ones.

In the context of our exercise, we found the discriminant \(D = 49\), a positive number, which means our quadratic equation has two distinct real solutions.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra. The steps to solve are sequential and methodical, starting with identifying the coefficients \(a\), \(b\), and \(c\) from the standard form of a quadratic equation \(ax^2 + bx + c = 0\). After identifying these coefficients, the next logical step is to calculate the discriminant using \(D = b^2 - 4ac\).
Once we have the discriminant, we can determine the nature of the solutions. If the discriminant is zero or positive, we proceed with the quadratic formula to find the real solutions. If it's negative, we might delve into complex numbers, but that's another topic. For our exercise, we're ensured two real solutions because we computed a positive discriminant, taking us forward to applying the quadratic formula for finding the precise values of those solutions.

Quadratic Formula

The quadratic formula, \(x = \frac{-b \pm \sqrt{D}}{2a}\), where \(D\) is the discriminant, is derived from completing the square of the general quadratic equation, and it provides a straightforward method for finding the real solutions of quadratic equations. What makes this formula so effective is its ability to systematically reveal the solutions, regardless of whether the discriminant indicates one or two real solutions. With our positive discriminant \(49\), plugging in the values of \(a\), \(b\), and \(c\) from \(-2k^2 + k + 6 = 0\) into the quadratic formula will yield the exact real solutions for \(k\).
It's important to catch that there are two possible solutions due to the \(\pm\) in the formula, which considers both the positive and negative square roots of the discriminant, ensuring we account for both potential intersection points of our quadratic equation with the x-axis when graphed.