Solving $2y(y-5) = 3(2y-1) - 4$ Expressing Solutions As Surds

In this article, we will delve into the process of solving the quadratic equation $2y(y-5) = 3(2y-1) - 4$, providing a comprehensive, step-by-step guide. Our primary goal is to find the solutions for y and express them in their simplest surd form. This involves manipulating the equation, simplifying it to its standard quadratic form, and then employing the quadratic formula to derive the solutions. Understanding the nuances of quadratic equations is fundamental in various fields of mathematics and practical applications, making this exercise not just an academic pursuit but also a valuable skill to acquire. By breaking down each step, we aim to make the solution accessible and clear, ensuring that anyone tackling similar problems can follow along with ease.

Understanding the Fundamentals of Quadratic Equations

Before diving into the specifics of our equation, let’s establish a firm grasp on the fundamentals of quadratic equations. A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term that is squared. The standard form of a quadratic equation is expressed as $ax^2 + bx + c = 0$, where a, b, and c are constants, and a is not equal to zero. The solutions to a quadratic equation, also known as roots or zeros, are the values of x that satisfy the equation. These solutions can be real or complex numbers, and a quadratic equation typically has two solutions, though they may be the same (a repeated root). The methods for solving quadratic equations include factoring, completing the square, and the quadratic formula. Each method has its advantages and is suited for different types of quadratic equations. For instance, factoring is efficient for equations that can be easily factored, while the quadratic formula is a universal method that can solve any quadratic equation, regardless of its complexity. In our case, we will use the quadratic formula to ensure accuracy and to demonstrate its application in detail.

Step 1: Expanding and Simplifying the Equation

Our initial equation is $2y(y-5) = 3(2y-1) - 4$. The first step in solving this equation is to expand and simplify both sides to bring it closer to the standard quadratic form. We begin by distributing the terms on both sides of the equation. On the left side, we distribute $2y$ across the parentheses, resulting in $2y^2 - 10y$. On the right side, we distribute the $3$ across the parentheses, yielding $6y - 3$. We then subtract $4$ from this result, giving us $6y - 7$. Now our equation looks like this: $2y^2 - 10y = 6y - 7$. To further simplify, we want to set the equation equal to zero, which is the standard form for a quadratic equation. We achieve this by moving all terms to one side of the equation. We subtract $6y$ from both sides, resulting in $2y^2 - 16y = -7$. Then, we add $7$ to both sides to get the equation in its standard form: $2y^2 - 16y + 7 = 0$. This simplified form is crucial because it allows us to easily identify the coefficients a, b, and c, which are necessary for applying the quadratic formula. In this case, we have $a = 2$, $b = -16$, and $c = 7$. With the equation in this form, we are now ready to proceed to the next step: applying the quadratic formula.

Step 2: Applying the Quadratic Formula

Now that we have the quadratic equation in the standard form $2y^2 - 16y + 7 = 0$, we can apply the quadratic formula to find the solutions for y. The quadratic formula is a powerful tool that provides a direct method for solving any quadratic equation of the form $ax^2 + bx + c = 0$. The formula is given by:

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

In our equation, we have $a = 2$, $b = -16$, and $c = 7$. Substituting these values into the quadratic formula, we get:

$$y = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(2)(7)}}{2(2)}$$

This substitution is a critical step, and it's important to ensure that the values are correctly placed in the formula. Next, we simplify the expression step by step. First, we calculate the value inside the square root. $(-16)^2$ equals $256$, and $4(2)(7)$ equals $56$. Thus, the expression inside the square root becomes $256 - 56$, which equals $200$. So now we have:

$$y = \frac{16 \pm \sqrt{200}}{4}$$

The next step is to simplify the square root term, $\sqrt{200}$. This involves finding the largest perfect square that divides 200. The largest perfect square factor of 200 is 100, since $200 = 100 \times 2$. Therefore, $\sqrt{200}$ can be written as $\sqrt{100 \times 2}$, which simplifies to $\sqrt{100} \times \sqrt{2}$, or $10\sqrt{2}$. Substituting this back into our equation, we now have:

$$y = \frac{16 \pm 10\sqrt{2}}{4}$$

We are now ready to simplify the entire expression, which we will tackle in the next step.

Step 3: Simplifying the Surd Form

Having reached the expression $y = \frac{16 \pm 10\sqrt{2}}{4}$, the final step is to simplify this surd form to its simplest expression. To do this, we look for common factors in the numerator that can be divided out with the denominator. Both 16 and $10\sqrt{2}$ are divisible by 2, and the denominator 4 is also divisible by 2. Therefore, we can factor out a 2 from the numerator:

$$y = \frac{2(8 \pm 5\sqrt{2})}{4}$$

Now, we can cancel the common factor of 2 between the numerator and the denominator. This simplifies the expression to:

$$y = \frac{8 \pm 5\sqrt{2}}{2}$$

This is the simplified surd form of the solutions. The "$\pm$" sign indicates that we have two distinct solutions, one with addition and one with subtraction. These solutions are:

$$y_1 = \frac{8 + 5\sqrt{2}}{2}$$

$$y_2 = \frac{8 - 5\sqrt{2}}{2}$$

These values are the solutions to the quadratic equation $2y(y-5) = 3(2y-1) - 4$, expressed in their simplest surd forms. We have successfully navigated the equation from its initial form, through expansion and simplification, application of the quadratic formula, and finally, simplification of the surd results. This process demonstrates the complete methodology for solving quadratic equations and expressing solutions in the required format. Understanding these steps is crucial for anyone studying algebra and related fields.

In summary, we have successfully solved the quadratic equation $2y(y-5) = 3(2y-1) - 4$ and expressed the solutions in their simplest surd form. The process involved several critical steps, including expanding and simplifying the equation to its standard quadratic form, applying the quadratic formula, and simplifying the resulting surd expressions. By meticulously following each step, we were able to arrive at the two solutions: $y_1 = \frac{8 + 5\sqrt{2}}{2}$ and $y_2 = \frac{8 - 5\sqrt{2}}{2}$. This exercise not only provides the specific solutions to this equation but also serves as a comprehensive guide for solving other quadratic equations. Mastering these techniques is essential for students and professionals alike, as quadratic equations frequently appear in various mathematical and real-world contexts. The ability to confidently solve these equations is a testament to a strong foundation in algebra and problem-solving skills. Furthermore, understanding the nature of solutions, whether they are rational, irrational, or complex, adds another layer of depth to mathematical proficiency. Quadratic equations are a cornerstone of mathematical education, and the principles learned through solving them extend to more advanced mathematical concepts.