Problem 82 - Intermediate Algebra

Question

Solve the system \(\left\\{\begin{array}{l}x^{2}-y^{2}=16 \\\ x^{2}+y^{2}=9\end{array}\right.\) over the complex numbers.

Step-by-Step Solution

Answer

The solutions are \((\pm \frac{5\sqrt{2}}{2}, \pm i\frac{\sqrt{7}}{\sqrt{2}})\).

Step 1: Understand the system

We have a system of equations: \( x^2 - y^2 = 16 \) and \( x^2 + y^2 = 9 \). The goal is to solve for \(x\) and \(y\) in the complex number system.

Step 2: Use algebraic identity

Recognize the difference of squares identity: \( (x^2 - y^2) = (x-y)(x+y)\). We have two equations: \( x^2 - y^2 = 16 \) and \( x^2 + y^2 = 9 \).

Step 3: Express in terms of squares

By adding and subtracting the given equations:\[ (x^2 + y^2) + (x^2 - y^2) = 9 + 16 \] which gives \( 2x^2 = 25 \) and \[ (x^2 + y^2) - (x^2 - y^2) = 9 - 16 \] which yields \( 2y^2 = -7 \).

Step 4: Solve for \(x^2\) and \(y^2\)

Divide the equations from Step 3:- \( x^2 = \frac{25}{2} \).- \( y^2 = -\frac{7}{2} \).

Step 5: Solve for \(x\) using complex numbers

Since \(x^2 = \frac{25}{2}\), take the square root:- \( x = \pm \sqrt{\frac{25}{2}} = \pm \frac{5\sqrt{2}}{2} \).

Step 6: Solve for \(y\) using complex numbers

Since \(y^2 = -\frac{7}{2}\), and it involves a negative square, use complex numbers:- \( y = \pm i\sqrt{\frac{7}{2}} = \pm i \frac{\sqrt{7}}{\sqrt{2}} \).

Step 7: Solutions for \(x\) and \(y\)

The solutions are:- \( x = \frac{5\sqrt{2}}{2} \), \( y = i\frac{\sqrt{7}}{\sqrt{2}} \)- \( x = -\frac{5\sqrt{2}}{2} \), \( y = -i\frac{\sqrt{7}}{\sqrt{2}} \)- \( x = \frac{5\sqrt{2}}{2} \), \( y = -i\frac{\sqrt{7}}{\sqrt{2}} \)- \( x = -\frac{5\sqrt{2}}{2} \), \( y = i\frac{\sqrt{7}}{\sqrt{2}} \)

Key Concepts

Understanding System of Equations

A system of equations is a set of two or more equations that have common variables. Solving these systems involves finding values for the variables that satisfy all equations in the system simultaneously. For the given exercise, we are dealing with two equations:

• \( x^2 - y^2 = 16 \)
• \( x^2 + y^2 = 9 \)

These equations form a relationship between the variables \( x \) and \( y \). To solve this system, we need to determine a set of \( x \) and \( y \) values that work for both equations at the same time. In this particular problem, we explore solutions over the complex numbers. This means that the solutions for \( x \) and \( y \) can include imaginary numbers, expanding our possibilities and revealing a broader spectrum of solutions. This can seem daunting at first, but with the right algebraic tools, it becomes quite manageable.

Using the Difference of Squares

The 'difference of squares' is a common algebraic identity used to simplify expressions and solve equations. It asserts that:
\((a^2 - b^2) = (a - b)(a + b)\).
In the given problem, the equation \( x^2 - y^2 = 16 \) can be seen through the lens of this identity:

• \((x - y)(x + y) = 16\)

Applying this identity can simplify the process of problem-solving by reducing the need for more complex arithmetic steps.Using it here, however, is primarily a step towards breaking the problem down. By interpreting the original equations this way, and using basic algebra, we can later deduce specific conditions that involve squaring both variables to eliminate one and focus inherently on another. This approach helps unravel the system into a more manageable form, leading us directly to useful numeric values.

Finding Complex Solutions

Complex solutions involve numbers that have both a real part and an imaginary part, expressed as \( a + bi \), where \( i \) is the imaginary unit \( \sqrt{-1} \). In this exercise, when we solve for \( y \), we encounter a negative square term:

• \( y^2 = -\frac{7}{2} \)

The negative sign under the square root indicates the nature of complex solutions. Within real numbers, you cannot take the square root of a negative number. However, in complex numbers, this is resolved by using the imaginary unit \( i \). Thus, this equation takes the following form as its solution:

• \( y = \pm i \sqrt{\frac{7}{2}} \)

Similarly, \( x \) is solved straightforwardly given its positive square:

• \( x = \pm \frac{5\sqrt{2}}{2} \)

These solutions highlight how complex numbers play a crucial role in extending the breadth of possible solutions, capturing scenarios real numbers alone cannot address. Understanding this expansion to complex numbers is key to solving many systems involving negative square roots effectively, allowing us to uncover all potential solutions.