Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the nonlinear system. $$ \begin{array}{c} x^{2}+y^{2}=6 \\ x y=1 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the System of Equations

We have two equations in the system: 1) \( x^2 + y^2 = 6 \) - a circle equation representing all points whose squared sum is 6, and 2) \( x y = 1 \) - a hyperbola equation representing the product of \( x \) and \( y \) equaling 1.

Step 2: Solve for One Variable from the Second Equation

From the equation \( x y = 1 \), express one variable in terms of the other. Let's solve for \( y \): \[ y = \frac{1}{x} \]

Step 3: Substitute the Expression into the First Equation

Substitute \( y = \frac{1}{x} \) into the first equation \( x^2 + y^2 = 6 \). This gives:\[ x^2 + \left(\frac{1}{x}\right)^2 = 6 \] Therefore:\[ x^2 + \frac{1}{x^2} = 6 \]

Step 4: Multiply Through by \( x^2 \) to Eliminate the Fraction

Multiply the entire equation by \( x^2 \) to eliminate the fraction:\[ x^4 + 1 = 6x^2 \] Rearrange into a standard quadratic equation in terms of \( x^2 \):\[ x^4 - 6x^2 + 1 = 0 \]

Step 5: Use a Substitution to Simplify the Equation

Let \( z = x^2 \), which means the equation becomes:\[ z^2 - 6z + 1 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -6, c = 1 \):\[ z = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]\[ z = \frac{6 \pm \sqrt{36 - 4}}{2} \]\[ z = \frac{6 \pm \sqrt{32}}{2} \]\[ z = \frac{6 \pm 4\sqrt{2}}{2} \]\[ z = 3 \pm 2\sqrt{2} \]

Step 7: Find the Values of \( x \)

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

Step 8: Find Corresponding Values of \( y \)

For each \( x \), use \( y = \frac{1}{x} \) to find \( y \):- If \( x = \sqrt{3 + 2\sqrt{2}} \), \( y = \frac{1}{\sqrt{3 + 2\sqrt{2}}} \)- If \( x = -\sqrt{3 + 2\sqrt{2}} \), \( y = -\frac{1}{\sqrt{3 + 2\sqrt{2}}} \)- If \( x = \sqrt{3 - 2\sqrt{2}} \), \( y = \frac{1}{\sqrt{3 - 2\sqrt{2}}} \)- If \( x = -\sqrt{3 - 2\sqrt{2}} \), \( y = -\frac{1}{\sqrt{3 - 2\sqrt{2}}} \).

Step 9: Write the Solution Pairs

So, the pairs \( (x, y) \) are: \((\sqrt{3 + 2\sqrt{2}}, \frac{1}{\sqrt{3 + 2\sqrt{2}}})\), \((-\sqrt{3 + 2\sqrt{2}}, -\frac{1}{\sqrt{3 + 2\sqrt{2}}})\), \((\sqrt{3 - 2\sqrt{2}}, \frac{1}{\sqrt{3 - 2\sqrt{2}}})\), \((-\sqrt{3 - 2\sqrt{2}}, -\frac{1}{\sqrt{3 - 2\sqrt{2}}})\).

Key Concepts

quadratic equations

Quadratic equations are polynomials of degree 2, which means the highest exponent of the variable is a square, or 2. These equations generally take the form:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants and \( x \) is the variable. We encounter quadratic equations often in problems involving areas, projectile motions, and here, in our nonlinear systems.
Quadratic equations are pivotal in solving nonlinear systems like the one posed in the exercise. In Step 4 of our solution, to solve the system \( x^4 - 6x^2 + 1 = 0 \), substituting \( z = x^2 \) transforms a quartic (degree 4) equation into a quadratic (degree 2) equation, simplifying the task significantly.
The quadratic formula is a powerful tool that helps us find the solutions of the quadratic equation:

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

This formula is derived from completing the square method, providing solutions to any quadratic equation assuming a real solution exists.

solutions of equations

Solving equations is about finding the values that satisfy the equations, making each equation a true statement. In the context of systems of equations, like our nonlinear system, we aim to find values that satisfy both equations simultaneously.
In the given exercise, the solutions are the pairs \((x, y)\) that meet both the circle equation \( x^2 + y^2 = 6 \) and the hyperbola equation \( xy = 1 \). Solving such systems requires ingenuity and often more than one technique. Here, substituting, rearranging, and applying algebraic tricks help us maintain clarity and order in solving these intricate problems.
After substitutions and transformations, finding the roots of quadratic equations yields potential solutions. Consideration of both positive and negative square roots is crucial since squaring a number results in a non-negative number. Therefore, solutions often occur in pairs, which we verify against the original equations for their validity.

substitution method

The substitution method is an algebraic technique used to solve systems of equations. It involves expressing one variable in terms of another from one equation and replacing (substituting) it in the other equation, eliminating one variable and allowing us to solve for the remaining unknown.
In our exercise, we used the substitution method starting in Step 2. By expressing \( y = \frac{1}{x} \) from the equation \( xy = 1 \), we substitute this expression into the circle equation \( x^2 + y^2 = 6 \). This substitution translates the problem into one that involves only one variable, making it easier to manipulate and solve.

• This particular substitution transforms the original problem into solving a single-variable polynomial equation.
• It emphasizes the power of substitution in reducing complexity, focusing on solving one variable at a time.
• It is a versatile method applicable to both linear and nonlinear systems.

By isolating and substituting variables methodically, the substitution method not only simplified computations but also paved the way for using other algebraic techniques effectively.