Problem 28 - Algebra and Trigonometry

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

From the equation \(xy = 24\), express \(y\) in terms of \(x\): \(y = \frac{24}{x}\). This allows us to substitute \(y\) in the second equation.

Step 2: Substitute y in the second equation

Use the expression for \(y\) from Step 1 in the second equation: \[2x^2 - \left(\frac{24}{x}\right)^2 + 4 = 0\]Simplify to find an equation solely in terms of \(x\).

Step 3: Simplify the equation

Simplifying the equation from Step 2, we have:\[2x^2 - \frac{576}{x^2} + 4 = 0\]Multiply through by \(x^2\) to eliminate the fraction:\[2x^4 + 4x^2 - 576 = 0\]

Step 4: Solve the quadratic equation

Introduce a substitution \(z = x^2\), which transforms the equation into a quadratic form: \[2z^2 + 4z - 576 = 0\]Solve this quadratic equation using the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 4\), and \(c = -576\).

Step 5: Compute solutions for z

Calculate:\[z = \frac{-4 \pm \sqrt{4^2 - 4 \times 2 \times (-576)}}{2 \times 2}\]\[z = \frac{-4 \pm \sqrt{16 + 4608}}{4}\]\[z = \frac{-4 \pm \sqrt{4624}}{4}\]\[z = \frac{-4 \pm 68}{4}\]This gives us \(z = 16\) or \(z = -18\).

Step 6: Find x from z values

Since \(z = x^2\), find \(x\):- For \(z = 16\), \(x^2 = 16\) implies \(x = 4\) or \(x = -4\).- The solution \(z = -18\) is not applicable for real \(x\) because \(x^2\) cannot be negative.

Step 7: Determine y values for each x

For each \(x\) value, find \(y\) using \(y = \frac{24}{x}\):- If \(x = 4\), then \(y = \frac{24}{4} = 6\).- If \(x = -4\), then \(y = \frac{24}{-4} = -6\).

Step 8: Conclude solutions

The system of equations has two solutions: \((x, y) = (4, 6)\) and \((x, y) = (-4, -6)\).

Key Concepts

Quadratic Equations

Quadratic equations are algebraic expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. These equations are called 'quadratic' because the highest power of \(x\) is 2, which means they form a parabola when graphed.
There are several methods to solve quadratic equations, including factorization, completing the square, and the quadratic formula. In the context of our system of equations, we transformed our original expression into a quadratic equation by using substitution, making it easier to solve using the quadratic formula.
Using these techniques allows us to find the roots or solutions of the quadratic equations, which are the values of \(x\) that make the equation equal to zero. Understanding this is crucial to solving any system of equations involving quadratics.

Substitution Method

The substitution method is a common technique used to solve a system of equations. This method involves expressing one variable in terms of another and then substituting this expression into the other equation. By doing so, the system is reduced to a single equation with one unknown variable.
In our exercise, we first isolated \(y\) in the equation \(xy = 24\) to obtain \(y = \frac{24}{x}\). We then substituted this expression into the second equation \(2x^2 - y^2 + 4 = 0\). This substitution reduces the equation to one variable, simplifying the problem significantly and allowing us to proceed with solving for \(x\).
By transforming the original system into a more manageable form, the substitution method makes solving complex problems more straightforward.

Real Solutions

Real solutions refer to the values of variables that satisfy an equation and are real numbers. In algebra, we deal primarily with real numbers, unless specified otherwise.
For quadratic equations, the discriminant \(b^2 - 4ac\) in the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) determines the nature of the solutions. If the discriminant is positive, there are two distinct real solutions. If it is zero, there is exactly one real solution (a repeated root). However, if it is negative, there are no real solutions, but instead, complex or imaginary solutions.

• Positive Discriminant: Two distinct real roots.
• Zero Discriminant: One real root.
• Negative Discriminant: No real roots, only complex numbers.

In our problem, solving the quadratic equation through substitution yielded \(z = 16\) and \(z = -18\). Since \(z = x^2\), only \(z = 16\), which equals \(x^2 = 16\), provided real solutions: \(x = 4\) and \(x = -4\). The solution \(z = -18\) was not viable because it did not provide real-number answers, highlighting the importance of understanding real solutions.

Algebraic Expressions

Algebraic expressions are mathematical phrases involving variables, constants, and arithmetic operations such as addition, subtraction, multiplication, and division. They form the foundation of algebra and are used to represent a wide range of mathematical and real-life problems.
In the given system of equations, both equations included algebraic expressions combining variables \(x\) and \(y\) with constants and operations.

• First Equation: \(xy = 24\) is a simple expression representing the product of \(x\) and \(y\).
• Second Equation: \(2x^2 - y^2 + 4 = 0\) is a more complex expression involving squares and a constant.

Manipulating these expressions is essential in solving systems of equations. It requires understanding how different operations affect the variables and how to solve for those variables. The process often involves rearranging, factoring, and simplifying these expressions, showcasing the essential role of algebra in problem-solving.