Problem 38 - College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Introduce the System of Equations

We have two equations with two variables:\[ x^2 + y^2 = 6 \]and \[ xy = 1. \]We need to find values of \(x\) and \(y\) that satisfy both equations.

Step 2: Express One Variable in Terms of the Other

From the second equation \(xy = 1\), we can express \(y\) in terms of \(x\) as \( y = \frac{1}{x} \). This allows us to substitute \(y\) in the first equation.

Step 3: Substitute and Simplify

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

Step 4: Multiply Equation by x²

To eliminate the fraction, multiply the entire equation by \(x^2\):\[x^4 + 1 = 6x^2.\]

Step 5: Rearrange into a Standard Form

Rearrange to obtain a polynomial equation:\[x^4 - 6x^2 + 1 = 0.\]

Step 6: Solve for x²

Let \(z = x^2\). The equation becomes a quadratic:\[z^2 - 6z + 1 = 0.\]Solve this using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 1\).

Step 7: Compute the Roots of z

Using the quadratic formula, find \(z\):\[z = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.\]Thus, \(z_1 = 3 + 2\sqrt{2}\) and \(z_2 = 3 - 2\sqrt{2}\).

Step 8: Find Possible x Values

Since \(z = x^2\), we have \(x^2 = 3 + 2\sqrt{2}\) or \(x^2 = 3 - 2\sqrt{2}\). Thus, the possible \(x\) values are:\[ x = \pm \sqrt{3 + 2\sqrt{2}}, \quad x = \pm \sqrt{3 - 2\sqrt{2}}. \]

Step 9: Determine Corresponding y Values

Recall that \(y = \frac{1}{x}\). So for each value of \(x\), compute \(y\).For example, if \(x = \sqrt{3 + 2\sqrt{2}}\), then \(y = \frac{1}{\sqrt{3 + 2\sqrt{2}}}\). Similarly, compute \(y\) for other values of \(x\).

Step 10: List All Solutions

Considering signs and simplifying square roots, the solution pairs \((x, y)\) are:1. \(\left(\sqrt{3 + 2\sqrt{2}}, \frac{1}{\sqrt{3 + 2\sqrt{2}}}\right)\)2. \(\left(-\sqrt{3 + 2\sqrt{2}}, -\frac{1}{\sqrt{3 + 2\sqrt{2}}}\right)\)3. \(\left(\sqrt{3 - 2\sqrt{2}}, \frac{1}{\sqrt{3 - 2\sqrt{2}}}\right)\)4. \(\left(-\sqrt{3 - 2\sqrt{2}}, -\frac{1}{\sqrt{3 - 2\sqrt{2}}}\right)\).

Key Concepts

Quadratic Formula

When solving certain polynomial equations, we often need to find the roots of quadratic equations. The quadratic formula is a powerful tool for this. Given an equation in the form \( ax^2 + bx + c = 0 \), you can solve it using the quadratic formula:

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

Here's how the formula works:- Calculate the discriminant, \( b^2 - 4ac \).
This shows whether the solutions are real or complex.
If the discriminant is positive, there are two distinct real solutions.
If zero, there's one real solution, referred to as a double root.
If negative, the solutions are complex.- Then apply the quadratic formula to find the roots.In the problem we are tackling, the substitution method transformed our original problem into a quadratic one, allowing us to use this formula. Understanding how to efficiently apply the quadratic formula not only helps solve equations but builds a foundation for advanced topics.

Substitution Method

The substitution method is an elegant way to solve systems of equations where you express one variable in terms of another. This tactic reduces the number of equations and makes complex systems more manageable:

• Start with one equation and solve it for one variable in terms of another.
• Substitute this expression into the remaining equations.
• Continue simplifying until you find the required variable values.

In our original problem, we did this by setting \( y = \frac{1}{x} \) from \( xy = 1 \). This substitution made the nonlinear system into a single polynomial equation \( x^2 + \frac{1}{x^2} = 6 \). While simplifying, keep a careful eye on restrictions, like when \( x eq 0 \), to avoid invalid solutions.

Polynomial Equations

Polynomial equations like \( x^4 - 6x^2 + 1 = 0 \) are equations that involve powers of a variable. Understanding them is essential, especially when working with systems that result in higher-order polynomials:

• The degree of the polynomial is the highest power of the variable, shaping its behavior like the number of solutions.
• Solutions to polynomial equations may be real or complex, and not all solutions are valid for given constraints.

Reducing the equation to a quadratic like \( z^2 - 6z + 1 = 0 \) makes it easier to apply the quadratic formula, as we did by setting \( z = x^2 \). For polynomial equations higher than second degree, special techniques or numerical methods may be used, but breaking them into smaller parts often helps clarify things and find the necessary solutions smoothly.