Problem 24 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 49 k^{2}+4=-28 k $$

Step-by-Step Solution

Answer

The solution to the equation is \(k = -\frac{2}{7}\).

Step 1: Write the Equation in Standard Form

The given equation is \(49k^2 + 4 = -28k\). To write it in standard form, we need to rearrange it as \(49k^2 + 28k + 4 = 0\). This form is \(ax^2 + bx + c = 0\) where \(a=49\), \(b=28\), and \(c=4\).

Step 2: Use the Quadratic Formula

To find the solution, apply the quadratic formula: \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 49\), \(b = 28\), \(c = 4\).

Step 3: Calculate the Discriminant

Calculate the discriminant \(\Delta = b^2 - 4ac\). Substitute \(b=28\), \(a=49\), \(c=4\): \(\Delta = 28^2 - 4 \times 49 \times 4 = 784 - 784 = 0\).

Step 4: Solve Using the Discriminant

Since the discriminant is 0, there is one real, repeated solution. Substitute \(a\), \(b\), and \(\Delta\) into the quadratic formula: \(k = \frac{-28 \pm \sqrt{0}}{2 \times 49} = \frac{-28}{98} = -\frac{2}{7}\).

Step 5: Verify the Solution

Substitute \(k = -\frac{2}{7}\) back into the original equation: \(49(-\frac{2}{7})^2 + 4 = -28(-\frac{2}{7})\). Simplify both sides: LHS = \(49(\frac{4}{49}) + 4 = 4 + 4 = 8\); RHS = \( +8\). Both sides are equal, confirming the solution.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool used to find the solutions of quadratic equations, which are any equations of the form \( ax^2 + bx + c = 0 \). It allows for calculating the roots without factoring.
The formula itself is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula is derived from completing the square method and serves as a straightforward mechanism to solve any quadratic equation, purely reliant on the coefficients \(a\), \(b\), and \(c\).
Applying this formula systematically involves computing the parts within it, especially the discriminant, and then solving for the variable, which in the given problem is \(k\). By substituting the values of \(a\), \(b\), and \(c\), we can assess not only the number but also the nature of the solutions.

Discriminant

The discriminant is a critical component of the quadratic formula and influences the nature of the roots. Calculated as \( \Delta = b^2 - 4ac \), it gives insights into the roots' nature:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), it results in one real, repeated solution, known as a double root.
• If \( \Delta < 0 \), the solutions are complex or imaginary.

In our specific exercise, the discriminant \( \Delta = 0 \), meaning that there is exactly one real solution for the variable \(k\). This determines that the quadratic equation has a perfect square trinomial, resulting in a double root situation.

Real Solutions

Real solutions are the values of the variable that satisfy the quadratic equation in real numbers. The discriminant provides valuable information about these solutions:

• When \( \Delta = 0 \), the quadratic formula yields a single real solution, as we noticed in the exercise. The solution appears as a repeated root.

For instance, in our problem, solving \(49k^2 + 28k + 4 = 0\) using the quadratic formula and the discriminant (\(\Delta = 0\)), we find that \(k = -\frac{2}{7}\) is the sole real solution. This solution has practical implications for scenarios represented by the quadratic equation, displaying circumstances or values that are achievable within the real number system.

Standard Form

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Arranging a quadratic equation in standard form is integral to solving it using methods like factoring, completing the square, or the quadratic formula.
In the initial step of working through our exercise, rearranging the given equation to \(49k^2 + 28k + 4 = 0\) is necessary. It expresses the quadratic last terms as constants and aligns the coefficients \(a\), \(b\), and \(c\) correctly.
Understanding the standard form not only supports calculation ease but also helps visualize the equation's components and their interplay, making it simpler to apply other mathematical procedures like graphing.