Problem 50 - College Algebra

Question

For the following exercises, find the solutions to the nonlinear equations with two variables. $$x^{2}-x y+y^{2}-2=0$$ $$\quad\quad\quad\quad x+3 y=4$$

Step-by-Step Solution

Answer

The solutions are \( (x_1, y_1) \) and \( (x_2, y_2) \) corresponding to the roots for \( y \).

Step 1: Identifying the System of Equations

We have two equations to solve:1. Nonlinear Equation: \( x^2 - xy + y^2 - 2 = 0 \) 2. Linear Equation: \( x + 3y = 4 \). Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solving the Linear Equation for One Variable

First, solve the linear equation \( x + 3y = 4 \) for \(x\):\[x = 4 - 3y\]

Step 3: Substituting into the Nonlinear Equation

Substitute \( x = 4 - 3y \) into the nonlinear equation \( x^2 - xy + y^2 - 2 = 0 \):\[(4 - 3y)^2 - (4 - 3y)y + y^2 - 2 = 0\]

Step 4: Expanding and Simplifying the Equation

Expand the terms in the substituted equation:\[(16 - 24y + 9y^2) - ((4 - 3y)y) + y^2 - 2 = 0\]Which simplifies to:\[16 - 24y + 9y^2 - 4y + 3y^2 + y^2 - 2 = 0\]

Step 5: Combining Like Terms

Combine like terms:\[13y^2 - 28y + 14 = 0\]

Step 6: Solving the Quadratic Equation

This is a standard quadratic equation \(13y^2 - 28y + 14 = 0\). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 13\), \(b = -28\), and \(c = 14\):\[y = \frac{28 \pm \sqrt{28^2 - 4 \times 13 \times 14}}{2 \times 13}\]

Step 7: Calculating the Discriminant and Roots

First calculate the discriminant:\[\sqrt{28^2 - 4 \times 13 \times 14} = \sqrt{784 - 728} = \sqrt{56} = 2\sqrt{14}\]So the roots are:\[y = \frac{28 \pm 2\sqrt{14}}{26}\]

Step 8: Simplifying the Roots for y

Simplifying the expression for \(y\), we get:\[y_1 = \frac{14 + \sqrt{14}}{13}\quad\text{and}\quad\y_2 = \frac{14 - \sqrt{14}}{13}\]

Step 9: Finding Corresponding x Values

Substitute \(y_1\) and \(y_2\) back into \(x = 4 - 3y\) to find corresponding \(x\) values:- For \(y_1\):\(x_1 = 4 - 3\left(\frac{14 + \sqrt{14}}{13}\right)\)- For \(y_2\):\(x_2 = 4 - 3\left(\frac{14 - \sqrt{14}}{13}\right)\).

Key Concepts

Linear Equation

Linear equations are mathematical expressions that form straight lines when plotted on a graph. They are called linear because of the use of the term 'line' in their characteristics. In these equations, both variables are raised to the power of one. They usually take the form: \[ax + by = c \] where:

• \(a\), \(b\) are constants (or coefficients).
• \(c\) is the constant term.
• \(x\) and \(y\) are variables.

For example, in the equation \(x + 3y = 4\), \(x\) and \(y\) represent the variables, while the numbers 1 and 3 are their coefficients, and 4 is the constant term. By solving linear equations, we find the specific values for \(x\) and \(y\) that make the equation true. To solve a linear equation, isolate one of the variables. In the provided example, you solve for \(x\) in terms of \(y\) to get \(x = 4 - 3y\). This expresses \(x\) purely in terms of \(y\), making it easier to substitute into another equation if needed, such as in systems of equations.

Quadratic Formula

The quadratic formula is a tool used to find the roots of quadratic equations, which are equations of the form: \[ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). Quadratic equations can have zero, one, or two solutions. The quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides the solutions for \(x\) by undoing the square in the equation. To use it, simply substitute the coefficients of your equation into the formula. The notation \(\pm\) means you will often get two solutions: one where you add the square root, and one where you subtract it. This dual solution property is what allows quadratic equations to have two distinct roots. In the solution to our system, after substituting \(x = 4 - 3y\) into the nonlinear equation, we arrive at the quadratic equation \(13y^2 - 28y + 14 = 0\). By applying the quadratic formula, you find the potential solutions for \(y\).

Discriminants

A discriminant is a part of the quadratic formula denoted as \(b^2 - 4ac\). It provides crucial information about the nature of the roots of a quadratic equation. The value of the discriminant tells us how many roots the quadratic equation will have and their nature:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If the discriminant is zero, there is exactly one real root, also called a repeated or double root.
• If the discriminant is negative, the quadratic equation has no real roots, but rather two complex (or imaginary) roots.

In solving the equation \(13y^2 - 28y + 14 = 0\), the discriminant is calculated as \(b^2 - 4ac = 28^2 - 4 \cdot 13 \cdot 14\). The result is \(56\), which is positive, indicating that our quadratic equation will have two distinct real roots. Once we know how many solutions there are, we can proceed to calculate them using the full quadratic formula. Understanding discriminants helps learners grasp why solutions to quadratic equations take the form they do, as well as foresee the nature of potential solutions.