Problem 38 - College Algebra

Question

Use any method to solve the nonlinear system. $$ \begin{array}{r} x^{2}+y^{2}=6 \\ x y=1 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \((\pm \sqrt{3+2\sqrt{2}}, \pm \frac{1}{\sqrt{3+2\sqrt{2}}})\) and \((\pm \sqrt{3-2\sqrt{2}}, \pm \frac{1}{\sqrt{3-2\sqrt{2}}})\).

Step 1: Express y in terms of x

Start with the second equation: \( xy = 1 \). Solve for \( y \) by dividing both sides by \( x \): \( y = \frac{1}{x} \).

Step 2: Substitute y in the first equation

Substitute \( y = \frac{1}{x} \) from Step 1 into the first equation \( x^2 + y^2 = 6 \). This results in the equation: \( x^2 + \left(\frac{1}{x}\right)^2 = 6 \).

Step 3: Simplify the equation

Rewrite the equation from Step 2 as: \( x^2 + \frac{1}{x^2} = 6 \). Multiply every term by \( x^2 \) to eliminate the fraction: \( x^4 + 1 = 6x^2 \).

Step 4: Rearrange and form a quadratic equation

Rearrange the terms: \( x^4 - 6x^2 + 1 = 0 \). Let \( z = x^2 \), so the equation becomes \( z^2 - 6z + 1 = 0 \).

Step 5: Solve the quadratic equation

Solve the quadratic equation \( z^2 - 6z + 1 = 0 \) using the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -6 \), \( c = 1 \). So, \( z = \frac{6 \pm \sqrt{36 - 4}}{2} \) which simplifies to \( z = \frac{6 \pm \sqrt{32}}{2} \).

Step 6: Simplify the roots of the quadratic equation

Simplify \( \sqrt{32} \) to \( 4\sqrt{2} \). Thus, the solutions for \( z \) are \( z = \frac{6 \pm 4\sqrt{2}}{2} \) or \( z = 3 \pm 2\sqrt{2} \).

Step 7: Find x for each z

Since \( z = x^2 \), solve for \( x \):\( x = \pm \sqrt{3 + 2\sqrt{2}} \) and \( x = \pm \sqrt{3 - 2\sqrt{2}} \).

Step 8: Calculate y corresponding to each x

For each value of \( x \), compute \( y = \frac{1}{x} \). This results in pairs of solutions: \((x, y) = \left( \sqrt{3 + 2\sqrt{2}}, \frac{1}{\sqrt{3 + 2\sqrt{2}}} \right)\), \((x, y) = \left( -\sqrt{3 + 2\sqrt{2}}, -\frac{1}{\sqrt{3 + 2\sqrt{2}}} \right)\), \((x, y) = \left( \sqrt{3 - 2\sqrt{2}}, \frac{1}{\sqrt{3 - 2\sqrt{2}}} \right)\), \((x, y) = \left( -\sqrt{3 - 2\sqrt{2}}, -\frac{1}{\sqrt{3 - 2\sqrt{2}}} \right)\).

Key Concepts

Quadratic Equation

The quadratic equation is a fundamental concept in algebra. It is a type of polynomial equation and is characterized by the highest power being two. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). At its core, this equation represents a parabola when graphically plotted on a coordinate plane. Quadratic equations are essential because they are the foundational building block for more complex problems.

To solve a quadratic equation like \( z^2 - 6z + 1 = 0 \), one of the most reliable methods is using the quadratic formula:

• \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula requires three components: coefficients \( a \), \( b \), and \( c \) which come directly from the equation's terms. Substituting these into the formula, and solving for \( z \), helps find the roots of the equation, which are the points where the parabola intersects the x-axis. Understanding how to apply and manipulate the quadratic formula is crucial for finding solutions to problems involving parabolic paths or areas.

Substitution Method

The substitution method is a strategic approach for solving systems of equations. It is particularly useful when handling nonlinear systems where at least one equation is not linear. The basic idea is to solve one of the equations for one variable, and then substitute that expression into the other equation.

For instance, in the system:

• \( xy = 1 \)
• \( x^2 + y^2 = 6 \)

By solving the first equation \( xy = 1 \) for \( y \), we find \( y = \frac{1}{x} \). Substituting this expression for \( y \) in the second equation allows us to simplify and work with only one variable (in this case, \( x \)). Transforming the original equations into a single-variable equation eases the complexity of solving the system because our task transforms from solving multiple equations to handling a single polynomial equation.

In our example, once substituted, we simplify the equation to remove fractions and work towards a format that is easier to solve using methods like completing the square, factoring, or employing the quadratic formula.

System of Equations

A system of equations is a collection of two or more equations sharing the same set of variables. Solving these systems is about finding values for the variables that satisfy all given equations simultaneously. Systems of equations are prevalent because they help model real-world scenarios where multiple conditions or constraints are in play.

In mathematics, systems can be categorized into linear and nonlinear systems based on the equations involved. A linear system involves only linear equations, where variables are raised to the power of one. A nonlinear system, as in our example, includes at least one equation that is not linear, often involving quadratic or higher-degree terms.

To tackle systems of equations, there are several methods available:

• **Graphical Method**: Plot both equations on a graph to find their intersection points.
• **Substitution Method**: Solve one equation for a variable and substitute it into the other equation.
• **Elimination Method**: Add or subtract equations to eliminate a variable, simplifying the system.

Each method has its advantages, and the choice depends on the system's complexity and the equations' types. The substitution method in particular is very effective for nonlinear systems like our exercise, where isolating one variable can help reduce the system to a single equation for easier solving.