Problem 37 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the nonlinear system. $$ \begin{array}{r} x^{2}+y^{2}-6 y=7 \\ x^{2}+y=1 \end{array} $$

Step-by-Step Solution

Answer

The system has no simple rational solution due to the irrational roots of \(x^2 = \frac{-5 \pm \sqrt{101}}{2}\).

Step 1: Identify the equations

We have two equations in the system:1. \(x^2 + y^2 - 6y = 7\)2. \(x^2 + y = 1\)

Step 2: Solve the second equation for y

From the second equation, solve for \(y\):\[x^2 + y = 1\] Subtract \(x^2\) from both sides to get:\[y = 1 - x^2\]

Step 3: Substitute y into the first equation

Substitute \(y = 1 - x^2\) from Step 2 into the first equation:\[x^2 + (1-x^2)^2 - 6(1-x^2) = 7\]

Step 4: Simplify and solve for x

Expand and simplify the equation:\[x^2 + (1 - 2x^2 + x^4) - 6 + 6x^2 = 7\]Combine like terms:\[x^4 + 5x^2 - 12 = 7\]Move 7 to the left side:\[x^4 + 5x^2 - 19 = 0\]

Step 5: Solve the quartic equation

Substitute \(z = x^2\), getting:\[z^2 + 5z - 19 = 0\]Solve this quadratic using the quadratic formula:\[z = \frac{-5 \pm \sqrt{5^2 + 4 \times 1 \times 19}}{2 \times 1}\]\[z = \frac{-5 \pm \sqrt{25 + 76}}{2}\]\[z = \frac{-5 \pm \sqrt{101}}{2}\]

Step 6: Solve for x using z

The values of \(z\) from Step 5 give us \(x^2 = z\). Since \(\sqrt{101}\) is not rational, find approximate values if needed, or state that there are no simple rational solutions for \(x\) and \(y\).

Step 7: Solve for y using x

Use \(y = 1 - x^2\) to find corresponding \(y\) values for each \(x\) found in Step 6.

Step 8: Conclude no explicit solution

Both \(x\) and \(y\) rely on irrational values derived from \(\sqrt{101}\). Therefore, an exact simple solution isn't feasible without a non-rational expression.

Key Concepts

Quadratic Substitution

Quadratic substitution is a strategy used in solving systems of equations, especially useful when dealing with nonlinear equations. In this method, you solve one of the equations for one of the variables, and then substitute the result into the other equation. This helps to reduce the system to one equation with a single variable, making it simpler to handle.
In our example, we had the two equations

• \( x^2 + y^2 - 6y = 7 \)
• \( x^2 + y = 1 \)

We solved the second equation for \( y \) by subtracting \( x^2 \) from both sides, giving \( y = 1 - x^2 \). This new expression for \( y \) was substituted into the first equation, transforming it from an equation in terms of both \( x \) and \( y \) to just \( x \).
This substitution not only helped streamline the problem but also made it possible to solve more complex systems when equations appear in different bases or degrees. It's a critical first step in reducing the complexity of nonlinear systems.

Quartic Equations

Quartic equations are polynomial equations of the fourth degree, which means they have the form \( ax^4 + bx^3 + cx^2 + dx + e = 0 \). Solving these equations can be quite challenging, often requiring specific techniques or substitutions to simplify the process.
In our exercise, after substituting \( y \) in the first equation and simplifying, we arrived at a quartic equation. It was expressed as \( x^4 + 5x^2 - 19 = 0 \). To simplify it, we used a technique called substitution, where we let \( z = x^2 \), effectively transforming our quartic into the quadratic \( z^2 + 5z - 19 = 0 \).
The significance of handling quartic equations this way lies in converting them into forms that are easier to solve, like quadratics, by using substitutions or factoring. This method leverages the relationships between the powers by breaking down complexities into more manageable pieces.

Irrational Solutions

Irrational solutions can arise in equations that do not neatly resolve into rational numbers. This happens when the solutions involve square roots of numbers that are not perfect squares, resulting in values that cannot be exactly expressed as simple fractions.
In the case of our system, solving the quadratic \( z^2 + 5z - 19 = 0 \) using the quadratic formula, we found that \( z = \frac{-5 \pm \sqrt{101}}{2} \). Since \( \sqrt{101} \) is not a rational number, these solutions are irrational.
Working with irrational solutions is important in capturing all potential outcomes of an equation. It highlights situations where approximations might be necessary for practical application because exact numeric expressions aren't feasible. In problem-solving, recognizing that irrational numbers are natural outcomes in many contexts can help students stay prepared for all solution types.