Solving 3x - 2√x - 5 = 0 Transforming Into Quadratic Form

In the realm of mathematics, recognizing and manipulating equations into familiar forms is a crucial skill. One such form is the quadratic form, which allows us to apply well-established techniques for solving quadratic equations to equations that may not initially appear quadratic. This article delves into the equation 3x - 2√x - 5 = 0, demonstrating how it can be expressed in quadratic form and exploring the substitution method used to achieve this transformation. We will dissect the process step-by-step, highlighting the underlying principles and providing a comprehensive understanding of this mathematical manipulation. The ability to recognize and convert equations into quadratic form opens up a wider range of problem-solving possibilities, making it a valuable tool in any mathematician's arsenal. This article serves as a guide to understanding this technique, offering a clear explanation and practical application for a deeper understanding of mathematical problem-solving.

Understanding Quadratic Form

To truly appreciate the transformation of the equation 3x - 2√x - 5 = 0 into quadratic form, we must first grasp the essence of a quadratic equation. A quadratic equation is a polynomial equation of the second degree, generally expressed in the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The key characteristic of a quadratic equation is the presence of a term with the variable raised to the power of two (x²). However, the concept of quadratic form extends beyond equations that strictly adhere to this format.

An equation is said to be in quadratic form if it can be manipulated into the standard quadratic equation structure by using a suitable substitution. This means that the equation, while not directly appearing as ax² + bx + c = 0, can be transformed into this form through a clever change of variable. The power of recognizing quadratic form lies in the fact that we can then employ the well-established methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula, to find the solutions to the original equation. This technique significantly expands our ability to tackle a wider variety of equations, making it a valuable asset in mathematical problem-solving. Recognizing the underlying structure and applying appropriate substitutions are key to unlocking the solutions hidden within these equations.

The process of transforming an equation into quadratic form often involves identifying a repeating expression within the equation. In the case of 3x - 2√x - 5 = 0, the expression √x appears alongside its square, which is x. This observation is crucial because it suggests a potential substitution that can simplify the equation into a quadratic form. By letting u = √x, we create a direct relationship between u and x that allows us to rewrite the original equation in terms of u. This substitution is the cornerstone of the transformation, effectively bridging the gap between the original equation and its quadratic form. Understanding this substitution process is paramount to mastering the technique of solving equations in quadratic form.

Transforming 3x - 2√x - 5 = 0 into Quadratic Form

Let's embark on the journey of transforming the equation 3x - 2√x - 5 = 0 into its quadratic form. The key to this transformation lies in recognizing the relationship between x and √x. We can observe that x is the square of √x, which hints at a strategic substitution. By introducing a new variable, we can effectively recast the equation in a more familiar quadratic structure.

The crucial step is to let u = √x. This substitution is the cornerstone of our transformation. Now, we need to express the original equation in terms of u. Since u = √x, it follows that u² = (√x)² = x. This relationship is paramount as it allows us to replace both x and √x in the original equation with expressions involving u. This substitution is not arbitrary; it is carefully chosen to exploit the inherent structure of the equation and pave the way for a quadratic form.

Substituting u² for x and u for √x in the original equation 3x - 2√x - 5 = 0, we obtain: 3(u²) - 2(u) - 5 = 0. This resulting equation, 3u² - 2u - 5 = 0, is a quadratic equation in the variable u. This is the quadratic form we sought. Notice how the substitution has effectively transformed the original equation, which may not have immediately appeared quadratic, into a standard quadratic equation. This transformation is the heart of the technique, allowing us to apply the tools and methods we know for solving quadratic equations to a broader class of problems. The ability to recognize such opportunities for substitution is a hallmark of mathematical proficiency, enabling us to simplify complex equations and find elegant solutions. Now that we have the equation in quadratic form, we can proceed to solve for u and subsequently find the solutions for x.

The Correct Substitution: u = √x

Having walked through the process of transforming the equation, the correct substitution becomes clear. As we established, the key to converting 3x - 2√x - 5 = 0 into quadratic form is to recognize the relationship between x and √x. The correct substitution, as demonstrated, is u = √x. This substitution allows us to rewrite the equation in terms of u, resulting in a standard quadratic equation that we can then solve using familiar techniques.

Let's briefly examine why the other options are incorrect. If we were to let u = x², u = x³, or u = ³√x, we would not be able to directly substitute and obtain a quadratic equation in the form au² + bu + c = 0. These substitutions would lead to more complex expressions that do not readily lend themselves to quadratic solution methods. The beauty of the substitution u = √x lies in its direct relationship to both x and √x within the original equation. This direct relationship is what allows for the seamless transformation into a quadratic form.

The correct substitution, u = √x, effectively simplifies the equation by creating a direct link between the terms. It transforms the original equation into 3u² - 2u - 5 = 0, which is a quadratic equation in the standard form. This transformation is crucial because it allows us to leverage our existing knowledge of quadratic equations to solve the original problem. The ability to identify the correct substitution is a critical skill in mathematics, and in this case, it unlocks the solution to an equation that might otherwise appear intractable. This underscores the importance of recognizing patterns and relationships within equations to guide our problem-solving strategies.

Solving the Quadratic Equation

Now that we've successfully transformed the original equation 3x - 2√x - 5 = 0 into the quadratic form 3u² - 2u - 5 = 0, the next step is to solve for u. There are several methods we can employ to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring provides a straightforward approach.

To factor the quadratic equation 3u² - 2u - 5 = 0, we need to find two numbers that multiply to (3)(-5) = -15 and add up to -2. These numbers are -5 and 3. We can then rewrite the middle term, -2u, as -5u + 3u, allowing us to factor by grouping:

3u² - 5u + 3u - 5 = 0 u(3u - 5) + 1(3u - 5) = 0 (3u - 5)(u + 1) = 0

Setting each factor equal to zero, we get two possible solutions for u:

3u - 5 = 0 => 3u = 5 => u = 5/3 u + 1 = 0 => u = -1

So, the solutions for u are u = 5/3 and u = -1. However, we're not quite finished. Remember, our original goal was to solve for x, not u. We used the substitution u = √x to simplify the equation, and now we need to reverse that substitution to find the values of x.

Finding the Solutions for x

Having solved for u, the final step is to determine the corresponding values of x. Recall our substitution: u = √x. To find x, we need to reverse this substitution by squaring both sides of the equation. This will give us x in terms of u. However, it's crucial to remember that squaring can sometimes introduce extraneous solutions, so we must verify our solutions at the end.

We have two values for u: u = 5/3 and u = -1. Let's substitute each of these back into the equation u = √x and solve for x.

For u = 5/3: 5/3 = √x Squaring both sides, we get (5/3)² = x, which simplifies to x = 25/9.

For u = -1: -1 = √x Squaring both sides, we get (-1)² = x, which simplifies to x = 1. However, it's important to recognize that the square root of a number cannot be negative. Therefore, u = -1 does not yield a valid solution for x in the original equation. This highlights the importance of checking for extraneous solutions when reversing a substitution that involves squaring.

Now, let's verify the solution x = 25/9 by substituting it back into the original equation 3x - 2√x - 5 = 0:

3(25/9) - 2√(25/9) - 5 = 0 (25/3) - 2(5/3) - 5 = 0 (25/3) - (10/3) - (15/3) = 0 0 = 0

The solution x = 25/9 satisfies the original equation. Therefore, the only valid solution for the equation 3x - 2√x - 5 = 0 is x = 25/9. This process demonstrates the importance of both the initial transformation into quadratic form and the subsequent verification of solutions to ensure accuracy and avoid extraneous results. The journey from the original equation to the final solution highlights the power of substitution and careful algebraic manipulation in solving complex mathematical problems.

In conclusion, we've successfully navigated the process of identifying and solving an equation in quadratic form. We began with the equation 3x - 2√x - 5 = 0, which, at first glance, doesn't appear to be a standard quadratic equation. However, by recognizing the relationship between x and √x, we were able to strategically apply the substitution u = √x. This substitution was the key to transforming the equation into the quadratic form 3u² - 2u - 5 = 0.

From there, we employed the factoring method to solve for u, obtaining the solutions u = 5/3 and u = -1. We then reversed the substitution to find the corresponding values of x. This process involved squaring both sides of the equation u = √x, which introduced the potential for extraneous solutions. We found that x = 25/9 was a valid solution, while x = 1, derived from u = -1, was extraneous because the square root of a number cannot be negative.

This exercise underscores the power of recognizing and utilizing quadratic form as a problem-solving technique. By transforming equations into a familiar structure, we can leverage well-established methods for finding solutions. Moreover, it highlights the critical importance of verifying solutions, especially when the solving process involves operations that can introduce extraneous results. The ability to manipulate equations, apply appropriate substitutions, and rigorously check solutions is a hallmark of mathematical proficiency. This understanding not only enhances our problem-solving capabilities but also deepens our appreciation for the elegance and interconnectedness of mathematical concepts. This technique can be applied to a wide range of equations, making it a valuable tool in any mathematical toolkit.