Evaluate X^2 + Y^2 If X - Y = 2 And Xy = 15

Introduction

In this article, we will delve into the problem of evaluating the expression $x^2 + y^2$, given the conditions that $x - y = 2$ and $xy = 15$. This is a classic algebraic problem that requires us to utilize our knowledge of algebraic identities and manipulation techniques. We will explore different approaches to solve this problem, ensuring a comprehensive understanding of the underlying concepts. Our main goal is to find the numerical value of $x^2 + y^2$ without explicitly solving for $x$ and $y$, which demonstrates a more efficient and elegant solution method. This approach emphasizes the power of algebraic identities in simplifying complex expressions and solving problems more effectively.

Understanding the Problem

Before we dive into the solution, let’s make sure we fully grasp the problem. We are given two equations:

1. $x - y = 2$
2. $xy = 15$

Our objective is to find the value of $x^2 + y^2$. A common mistake is to try to solve for $x$ and $y$ individually and then substitute those values into the expression $x^2 + y^2$. While this approach will eventually yield the correct answer, it can be more time-consuming and prone to errors. Instead, we will use algebraic identities to directly find the value of $x^2 + y^2$ without explicitly solving for $x$ and $y$. This method not only simplifies the process but also highlights the importance of recognizing and applying appropriate algebraic identities. The key is to identify a relationship between the given equations and the expression we want to evaluate.

Method 1: Using the Identity $(x - y)^2 = x^2 - 2xy + y^2$

One of the most straightforward approaches involves using the algebraic identity:

$(x - y)^2 = x^2 - 2xy + y^2$

We can rearrange this identity to isolate the term we are interested in, which is $x^2 + y^2$:

$x^2 + y^2 = (x - y)^2 + 2xy$

Now, we can substitute the given values of $x - y = 2$ and $xy = 15$ into this equation:

$x^2 + y^2 = (2)^2 + 2(15)$

$x^2 + y^2 = 4 + 30$

$x^2 + y^2 = 34$

Therefore, the value of $x^2 + y^2$ is 34. This method is efficient because it directly utilizes the given information and avoids the need to solve for $x$ and $y$ individually. The algebraic identity serves as a bridge connecting the given conditions with the expression we need to evaluate, making the solution process concise and clear.

Method 2: Solving for $x$ and $y$ and Substituting

While we aim to avoid solving for $x$ and $y$ directly, it's beneficial to illustrate how this method works for comparison and to provide a complete understanding of the problem. From the equation $x - y = 2$, we can express $x$ in terms of $y$:

$x = y + 2$

Substitute this expression for $x$ into the second equation, $xy = 15$:

$(y + 2)y = 15$

$y^2 + 2y = 15$

$y^2 + 2y - 15 = 0$

Now, we have a quadratic equation in terms of $y$. We can factor this equation:

$(y + 5)(y - 3) = 0$

This gives us two possible values for $y$:

$y = -5$ or $y = 3$

For each value of $y$, we can find the corresponding value of $x$ using $x = y + 2$:

If $y = -5$, then $x = -5 + 2 = -3$

If $y = 3$, then $x = 3 + 2 = 5$

Now we have two pairs of solutions: $(-3, -5)$ and $(5, 3)$. Let's substitute each pair into the expression $x^2 + y^2$:

For $(-3, -5)$:

$x^2 + y^2 = (-3)^2 + (-5)^2 = 9 + 25 = 34$

For $(5, 3)$:

$x^2 + y^2 = (5)^2 + (3)^2 = 25 + 9 = 34$

In both cases, we get $x^2 + y^2 = 34$. This method, while more involved, confirms our result from Method 1. It also illustrates that there can be multiple solutions for $x$ and $y$ that satisfy the given conditions, but the value of $x^2 + y^2$ remains the same. This approach highlights the importance of considering all possible solutions when dealing with systems of equations.

Comparison of Methods

As we have seen, both methods lead to the same answer, $x^2 + y^2 = 34$. However, Method 1, which utilizes the algebraic identity $(x - y)^2 = x^2 - 2xy + y^2$, is significantly more efficient. It avoids the need to solve a quadratic equation and directly computes the value of $x^2 + y^2$ by substituting the given values. This method demonstrates the power of algebraic manipulation and the strategic use of identities to simplify problems.

Method 2, on the other hand, involves solving for $x$ and $y$ individually, which requires factoring a quadratic equation and substituting the resulting values back into the expression. While this method is more laborious, it provides a comprehensive understanding of the possible solutions for $x$ and $y$. It also serves as a good check for the result obtained in Method 1.

In general, when faced with similar problems, it is often more efficient to look for algebraic identities that can directly relate the given information to the expression we want to evaluate. This approach not only saves time but also reduces the chances of making errors in the calculations.

Key Takeaways

• Algebraic Identities: The identity $(x - y)^2 = x^2 - 2xy + y^2$ is a powerful tool for solving problems involving the difference and product of two variables.
• Efficient Problem Solving: Using algebraic identities can often lead to more efficient solutions compared to solving for variables individually.
• Understanding Relationships: Recognizing the relationships between given equations and the expression to be evaluated is crucial for problem-solving.
• Multiple Solution Paths: There can be multiple ways to solve a problem, but some methods are more efficient than others.

Conclusion

In conclusion, we have successfully evaluated $x^2 + y^2$ given that $x - y = 2$ and $xy = 15$. We demonstrated two methods: one using the algebraic identity $(x - y)^2 = x^2 - 2xy + y^2$ and the other by solving for $x$ and $y$ directly. The first method proved to be more efficient, highlighting the importance of utilizing algebraic identities. The value of $x^2 + y^2$ was found to be 34. This problem underscores the significance of algebraic manipulation and strategic problem-solving techniques in mathematics. By understanding and applying these concepts, we can tackle complex problems with greater ease and efficiency. Remember, the key to solving such problems often lies in recognizing the underlying relationships and choosing the most appropriate method to leverage them.