Problem 24 - Precalculus: Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x y &=24 \\ 2 x^{2}-y^{2}+4 &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((4, 6)\) and \((-4, -6)\).

Step 1: Express y in Terms of x

Start with the first equation: \( x y = 24 \). To express \( y \) in terms of \( x \), solve for \( y \): \( y = \frac{24}{x} \).

Step 2: Substitute y in the Second Equation

Take the expression \( y = \frac{24}{x} \) and substitute it into the second equation \( 2x^2 - y^2 + 4 = 0 \). This gives us: \( 2x^2 - \left( \frac{24}{x} \right)^2 + 4 = 0 \).

Step 3: Simplify the Equation

Simplify the equation from Step 2: \( 2x^2 - \frac{576}{x^2} + 4 = 0 \). Multiply through by \( x^2 \) to eliminate the fraction: \( 2x^4 + 4x^2 - 576 = 0 \).

Step 4: Solve the Polynomial

Let \( z = x^2 \), then the equation is \( 2z^2 + 4z - 576 = 0 \). Divide the entire equation by 2: \( z^2 + 2z - 288 = 0 \). Factor to find: \( (z - 16)(z + 18) = 0 \). Thus, \( z = 16 \) or \( z = -18 \).

Step 5: Solve for x

Since \( z = x^2 \), then for \( z = 16 \), we have \( x^2 = 16 \), giving \( x = 4 \) or \( x = -4 \). The solution \( z = -18 \) is not valid as \( x^2 \) cannot be negative.

Step 6: Find Corresponding y Values

Use the valid \( x \) values to find \( y \). For \( x = 4 \), \( y = \frac{24}{4} = 6 \). For \( x = -4 \), \( y = \frac{24}{-4} = -6 \).

Step 7: Write the Solution

The solutions for the system are \((x, y) = (4, 6)\) and \((x, y) = (-4, -6)\).

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions involving variables and coefficients, where the variables are raised to whole number powers. They are a fundamental part of algebra and appear frequently in many areas of mathematics. In this particular problem, we encountered a polynomial equation derived from a system of equations. The original equations were:

• \( xy = 24 \)
• \( 2x^2 - y^2 + 4 = 0 \)

We transformed these into a polynomial equation, \( 2x^4 + 4x^2 - 576 = 0 \), which is a fourth-degree polynomial because the highest power of \( x \) is 4.
This equation was then expressed in terms of a new variable \( z = x^2 \), making it easier to handle as a quadratic equation, \( z^2 + 2z - 288 = 0 \). Polynomial equations like these can often be tackled by substitution, factoring, or using various algebraic methods, which simplifies the process of finding solutions.

Solving Equations

Solving equations is the process of finding the unknown values that make the equation true. In this problem, we solved a system of equations, which are two or more equations that share variables. To solve this system, we first expressed one variable in terms of the other using the given equations. Then, we substituted this expression into the second equation. This method turned the problem into a single equation with one variable, which is easier to solve. Here, after substituting \( y = \frac{24}{x} \) into the second equation, we obtained a polynomial equation.
Solving this involved using algebraic techniques, starting by simplifying the equation and then factorizing the quadratic equation. This approach is systematic and helps in breaking down complex equations into manageable parts. Finally, the solutions to the individual equations are used to find solutions to the entire system.

Algebraic Manipulation

Algebraic manipulation is a crucial skill that involves rearranging and simplifying equations to make them more solvable. In the provided problem, we engaged in several manipulations:

• We expressed \( y \) in terms of \( x \) by rearranging the first equation, \( xy = 24 \), to get \( y = \frac{24}{x} \).
• Next, we substituted this into the second equation, transforming it into a single-variable polynomial equation.
• To eliminate fractions, we multiplied through by \( x^2 \), simplifying the expression considerably.

Then, by letting \( z = x^2 \), we further reduced the equation to a familiar quadratic form, \( z^2 + 2z - 288 = 0 \).
These steps demonstrate the power of algebraic manipulation in solving complex mathematical tasks. Techniques like substitution, factoring, and rearranging are foundational tools for simplifying and solving difficult problems efficiently.