Solving $-2y^2 - 5y + 6 = 0$ With Quadratic Formula

Introduction to the Quadratic Formula

When we encounter a quadratic equation in the form of $ax^2 + bx + c = 0$, one of the most reliable methods to find the solutions (also known as roots) is by using the quadratic formula. The quadratic formula is a powerful tool that provides a straightforward way to solve for the variable, $x$, regardless of the complexity of the coefficients $a$, $b$, and $c$. This method is especially useful when factoring the quadratic equation is not immediately obvious or proves to be challenging. Understanding and applying the quadratic formula is a fundamental skill in algebra, and it opens doors to solving a wide range of mathematical problems in various fields, including physics, engineering, and economics.

The quadratic formula is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this comprehensive guide, we will walk through the process of using the quadratic formula to solve the specific quadratic equation $-2y^2 - 5y + 6 = 0$. We'll break down each step, ensuring a clear understanding of how to apply the formula effectively. This example will not only help you solve this particular equation but also provide you with a solid foundation for tackling other quadratic equations in the future. Whether you're a student learning algebra or someone looking to refresh your math skills, this guide will provide you with the knowledge and confidence to master the quadratic formula.

Identifying Coefficients

The first crucial step in solving a quadratic equation using the quadratic formula is to correctly identify the coefficients $a$, $b$, and $c$. These coefficients are the numerical values that correspond to the terms in the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. Accurate identification of these coefficients is paramount because they are directly substituted into the quadratic formula. A mistake in identifying even one coefficient can lead to an incorrect solution. Therefore, careful attention to detail is essential at this stage.

In our specific equation, $-2y^2 - 5y + 6 = 0$, we need to map each term to its corresponding coefficient. The term $-2y^2$ corresponds to the $ax^2$ term in the standard form, which means that the coefficient $a$ is -2. Similarly, the term $-5y$ corresponds to the $bx$ term, indicating that the coefficient $b$ is -5. Finally, the constant term +6 corresponds to the $c$ term, so the coefficient $c$ is 6. It's important to note the signs of the coefficients; a negative sign must be included, as seen with $a = -2$ and $b = -5$. Overlooking the negative signs is a common error that can be easily avoided with careful attention.

To summarize, we have:

• $a = -2$
• $b = -5$
• $c = 6$

With these coefficients correctly identified, we are now well-prepared to substitute them into the quadratic formula and proceed with solving the equation. This meticulous approach to identifying coefficients is the cornerstone of successfully applying the quadratic formula and finding the correct solutions.

Applying the Quadratic Formula

Now that we have identified the coefficients $a = -2$, $b = -5$, and $c = 6$, the next step is to substitute these values into the quadratic formula. The quadratic formula, as we recall, is given by:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Carefully substituting the values, we get:

$$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-2)(6)}}{2(-2)}$$

The importance of paying close attention to signs cannot be overstated here. The negative signs in the formula and the coefficients must be handled meticulously to avoid errors. For instance, $-(-5)$ becomes positive 5, and the product of negative numbers must be calculated correctly. This step is a critical juncture in the problem-solving process, where a simple mistake can lead to a completely different answer. Therefore, double-checking each substitution and calculation is a good practice to ensure accuracy.

After substitution, the formula looks like this:

$$y = \frac{5 \pm \sqrt{25 + 48}}{-4}$$

We have simplified the initial expression by addressing the negative signs and performing the basic multiplications and squares. The next step involves simplifying the expression under the square root, which is a crucial part of solving the equation. This careful and methodical approach to substitution and initial simplification sets the stage for a smooth and accurate solution using the quadratic formula. The goal is to break down the complex formula into manageable parts, making the process less daunting and more accessible.

Simplifying the Expression

Following the substitution of the coefficients into the quadratic formula, the next crucial step is to simplify the expression. This involves focusing on the part under the square root, which is known as the discriminant, and then further simplifying the entire expression. The discriminant, $b^2 - 4ac$, provides valuable information about the nature of the roots of the quadratic equation, but for now, our primary focus is on calculating its value accurately.

From our previous step, we have:

$$y = \frac{5 \pm \sqrt{25 + 48}}{-4}$$

First, we simplify the expression inside the square root. We have $25 + 48$, which equals 73. So the equation becomes:

$$y = \frac{5 \pm \sqrt{73}}{-4}$$

At this point, we have simplified the expression as much as possible without calculating the square root of 73. Since 73 is a prime number, its square root cannot be simplified to a whole number or a simple fraction. Therefore, we leave it as $\sqrt{73}$. The $\pm$ sign in the formula indicates that we have two possible solutions for $y$, one where we add the square root of 73 and one where we subtract it. This is a characteristic feature of quadratic equations, which often have two distinct solutions.

Simplifying the expression is a critical step because it reduces the complexity of the equation and makes it easier to handle. By accurately calculating the discriminant and simplifying the square root (if possible), we are setting ourselves up for the final step of finding the two solutions for $y$. The care and precision taken in this simplification process are essential for arriving at the correct answer. Now, we proceed to separate the two possible solutions and calculate them individually.

Calculating the Two Solutions

After simplifying the expression, we arrive at a point where we have two potential solutions for $y$, stemming from the $\pm$ sign in the quadratic formula. This sign signifies that we need to consider both the addition and subtraction of the square root term to find the two roots of the quadratic equation. Separating these two possibilities is the key to obtaining the final solutions.

From our previous simplification, we have:

$$y = \frac{5 \pm \sqrt{73}}{-4}$$

This gives us two equations:

1. $$y_1 = \frac{5 + \sqrt{73}}{-4}$$

2. $$y_2 = \frac{5 - \sqrt{73}}{-4}$$

To find the approximate values of these solutions, we need to calculate the square root of 73. The square root of 73 is approximately 8.544. Now we can substitute this value into our equations:

1. $$y_1 = \frac{5 + 8.544}{-4} = \frac{13.544}{-4} \approx -3.386$$

2. $$y_2 = \frac{5 - 8.544}{-4} = \frac{-3.544}{-4} \approx 0.886$$

Therefore, the two solutions for the quadratic equation $-2y^2 - 5y + 6 = 0$ are approximately $y_1 \approx -3.386$ and $y_2 \approx 0.886$. These are the values of $y$ that, when substituted back into the original equation, will make the equation true. Calculating these two solutions is the final step in solving the quadratic equation using the quadratic formula. It's a culmination of all the previous steps, from identifying coefficients to simplifying the expression, and it provides us with the answer we were seeking.

Conclusion

In this comprehensive guide, we have successfully navigated the process of solving the quadratic equation $-2y^2 - 5y + 6 = 0$ using the quadratic formula. We began by understanding the quadratic formula itself, which is a fundamental tool for solving equations of the form $ax^2 + bx + c = 0$. We then moved on to the critical step of identifying the coefficients $a$, $b$, and $c$ from the given equation, emphasizing the importance of careful attention to signs.

Next, we demonstrated how to substitute these coefficients into the quadratic formula, highlighting the need for precision to avoid errors. We then simplified the expression, focusing on the discriminant ($b^2 - 4ac$) and simplifying the square root term. This simplification allowed us to clearly see the two potential solutions arising from the $\pm$ sign in the formula. Finally, we calculated these two solutions, providing approximate values for the roots of the equation.

Through this step-by-step approach, we have shown that the quadratic formula is a powerful and reliable method for solving quadratic equations. While the formula may seem daunting at first, breaking it down into smaller, manageable steps makes the process much more accessible. This guide not only provides the solution to the specific equation but also equips you with the skills and understanding to tackle other quadratic equations with confidence. Mastering the quadratic formula is an invaluable asset in mathematics and opens doors to solving a wide array of problems in various fields.

For further exploration and practice with quadratic equations and the quadratic formula, you can visit resources like Khan Academy's Quadratic Equations Section. This can provide additional examples and explanations to reinforce your understanding.