Real Solutions Of A Quadratic Equation: A Step-by-Step Guide

Hey guys! Let's dive into a common type of math problem: figuring out how many real solutions a quadratic equation has. Today, we're tackling the equation $-15x^2 - 3 = 2(7x^2 - 1.5)$. This might look intimidating at first, but don't worry, we'll break it down step by step. Our main goal here is to determine just how many real number solutions this equation actually possesses. We aren’t necessarily trying to find the solutions themselves right away, but rather to learn about the nature and number of solutions it has. Let's get started and make math a little less scary, one equation at a time.

Understanding the Problem

First, let's understand what the question is asking. When we talk about "real solutions," we're referring to the values of x that, when plugged into the equation, make the equation true and are real numbers (not imaginary numbers). Quadratic equations, which are equations in the form of $ax^2 + bx + c = 0$, can have two real solutions, one real solution (which we sometimes call a repeated or double root), or no real solutions. The number of real solutions is closely linked to the famous discriminant, which we will discuss later. Knowing what constitutes a real solution is pivotal to understanding how we approach the problem at hand. Our aim is to transform and simplify the given equation to a standard quadratic form so we can identify the coefficients, and then use these coefficients to determine the number of solutions.

The key to solving this problem is to manipulate the given equation into a standard quadratic form. This form allows us to easily identify the coefficients a, b, and c, which are crucial for determining the number of real solutions. Remember, a quadratic equation is generally expressed as $ax^2 + bx + c = 0$, where a, b, and c are constants, and x is the variable we're solving for. By rearranging and simplifying our initial equation, we can fit it into this format, which will make our next steps much clearer. We need to handle algebraic manipulations like distribution, combining like terms, and rearranging the terms, all of which are important steps in simplifying and solving equations. Mastering these algebraic techniques is fundamentally important not just for this problem, but for numerous algebraic problems you'll encounter in mathematics.

So, why bother putting it in standard form? Well, the standard form of a quadratic equation is super helpful because it directly leads us to a powerful tool called the discriminant. The discriminant, which we'll explore in more detail later, is a part of the quadratic formula that tells us exactly how many real solutions our equation has without us having to fully solve for those solutions. It's a bit like a shortcut that saves us time and effort! Understanding the significance of the standard form helps underscore why we manipulate equations in algebra. It’s not just about rearranging symbols; it’s about putting the equation into a form that unlocks powerful analytical tools and insights. Now, let's move on to the next part and start actually simplifying the equation!

Simplifying the Equation

Alright, let's get our hands dirty and simplify the equation $-15x^2 - 3 = 2(7x^2 - 1.5)$. The first thing we need to do is get rid of those parentheses. We'll distribute the 2 on the right side of the equation. Remember, distributing means multiplying the number outside the parentheses by each term inside. So, $2 * 7x^2$ gives us $14x^2$, and $2 * -1.5$ gives us -3. Therefore, after distributing, our equation looks like this: $-15x^2 - 3 = 14x^2 - 3$. This step is all about clearing the way for further simplification by dealing with any immediate groupings. Distribution is a cornerstone technique in algebra and is frequently employed across a variety of mathematical contexts.

Now that we've distributed, it's time to gather like terms. Like terms are those that contain the same variable raised to the same power. In our equation, we have $x^2$ terms and constant terms. Let's move all the $x^2$ terms to one side of the equation and the constants to the other. A smart move here is to add $15x^2$ to both sides of the equation. This will eliminate the $-15x^2$ on the left side and keep our $x^2$ terms positive, which can make things easier later on. Doing this gives us $-3 = 14x^2 + 15x^2 - 3$. We now combine like terms on the right side to get $-3 = 29x^2 - 3$. This step highlights the strategy of isolating variables in equation solving. Collecting like terms allows us to consolidate and simplify the expression, bringing us closer to the standard quadratic form. It is a vital step in solving equations and understanding how different terms interact with each other.

We're almost there! To further isolate the $x^2$ term, we can add 3 to both sides of the equation. This eliminates the constant term on the right side, leaving us with $0 = 29x^2$. Now, we have a much simpler equation to work with. This simplification step further illustrates the principle of maintaining balance in equations. By performing the same operation on both sides, we preserve the equality while driving towards a more manageable form. In this particular case, the equation has simplified to such an extent that we can now easily assess the number of solutions.

Determining the Number of Real Solutions

We've simplified the equation down to $0 = 29x^2$. Now, let's think about what this means. To find the solutions, we need to solve for x. We can divide both sides of the equation by 29, which gives us $0 = x^2$. So, we're asking ourselves, "What number, when squared, equals zero?" Well, the only number that fits that description is 0 itself! Therefore, $x = 0$ is the only solution to this equation. This step is crucial in showing how simplifying an equation can lead to a direct and clear solution. It emphasizes the significance of simplification as a strategic tool in mathematics, especially when dealing with equations that initially seem complex.

But hold on, we need to be a bit careful here. While we found that x = 0 is a solution, we need to determine how many real solutions there are. In this case, we have one real solution, which is 0. However, it's considered a repeated solution (or a double root) because in the context of quadratic equations, a single root solution means that the graph of the corresponding quadratic equation touches the x-axis at just one point (in this case, the origin). It’s a fine but important distinction to make, as it ties the algebraic solution back to its geometric interpretation. Understanding the concept of repeated roots is vital for a more profound comprehension of quadratic behavior.

Now, let’s connect this to the discriminant, which I mentioned earlier. Remember, the discriminant is the part of the quadratic formula under the square root sign, $b^2 - 4ac$. It tells us the nature and number of solutions without having to solve the entire quadratic formula. In our original equation, after simplifying to the standard form, we can think of our equation as $29x^2 + 0x + 0 = 0$. So, a = 29, b = 0, and c = 0. Plugging these values into the discriminant, we get $0^2 - 4 * 29 * 0 = 0$. When the discriminant is equal to zero, it indicates that the quadratic equation has exactly one real solution (a repeated root), which perfectly matches our result. Linking the concept of the discriminant to the solution obtained reinforces its importance as a tool for determining the nature of roots in quadratic equations. It ties together theoretical understanding with practical problem-solving, making the solution process more coherent and meaningful.

Conclusion

So, guys, after simplifying the equation $-15x^2 - 3 = 2(7x^2 - 1.5)$ and analyzing it, we found that it has exactly one real solution. This solution is x = 0, and it's considered a repeated root. We also saw how the discriminant, $b^2 - 4ac$, can help us determine the number of real solutions without fully solving the equation. In this case, the discriminant was 0, confirming our finding of one real solution. This underscores the idea that mathematics provides multiple paths to the same answer, each offering a different perspective and insight. Understanding these varied approaches enriches problem-solving skills and cultivates a deeper appreciation for the elegance and interconnectedness of mathematics.

I hope this step-by-step explanation made things clearer for you. Remember, practice makes perfect, so keep tackling those quadratic equations! Understanding how to determine the number of real solutions is a fundamental skill in algebra and will serve you well in more advanced math courses. It’s also a great example of how a structured, step-by-step approach can turn seemingly complex problems into manageable tasks. By mastering techniques like simplifying equations, understanding the discriminant, and relating algebraic solutions to their geometric interpretations, you will be well-equipped to tackle a wide range of mathematical challenges. Keep practicing, stay curious, and math will become less like a challenge and more like an engaging puzzle. Until next time, keep solving!