Problem 52 - College Algebra with Corequisite Support

Question

For the following exercises, find the solutions to the nonlinear equations with two variables. $$ \begin{array}{l} x^{2}+4 x y-2 y^{2}-6=0 \\ x=y+2 \end{array} $$

Step-by-Step Solution

Answer

Solutions: \((-4 + \sqrt{42}/3, -6 + \sqrt{42}/3)\) and \((-4 - \sqrt{42}/3, -6 - \sqrt{42}/3)\).

Step 1: Substitute the Second Equation into the First

Given the equations: \(x^2 + 4xy - 2y^2 - 6 = 0\) and \(x = y + 2\), substitute \(x = y + 2\) into the first equation. This yields \((y + 2)^2 + 4(y + 2)y - 2y^2 - 6 = 0\).

Step 2: Expand and Simplify

Expand \((y + 2)^2\) to get \(y^2 + 4y + 4\). Expand \(4(y + 2)y\) to get \(4y^2 + 8y\). Substitute these back into the equation: \( y^2 + 4y + 4 + 4y^2 + 8y - 2y^2 - 6 = 0\). Simplify to \(3y^2 + 12y - 2 = 0\).

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \(3y^2 + 12y - 2 = 0\) using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a = 3\), \(b = 12\), \(c = -2\). Calculate the discriminant \(b^2 - 4ac = 144 + 24 = 168\). So, \(y = \frac{-12 \pm \sqrt{168}}{6}\). Simplify to find \(y = -6 \pm \sqrt{42}/3\).

Step 4: Find Corresponding x-values

Using \(x = y + 2\), calculate the \(x\)-values for each \(y\). For \(y_1 = -6 + \sqrt{42}/3\), \(x_1 = -4 + \sqrt{42}/3\). For \(y_2 = -6 - \sqrt{42}/3\), \(x_2 = -4 - \sqrt{42}/3\).

Step 5: Present the Solutions

The solutions for the nonlinear equations are: \((x_1, y_1) = (-4 + \sqrt{42}/3, -6 + \sqrt{42}/3)\) and \((x_2, y_2) = (-4 - \sqrt{42}/3, -6 - \sqrt{42}/3)\).

Key Concepts

Quadratic Equation

In mathematics, a quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( x \) represents an unknown variable, and \( a \), \( b \), and \( c \) are coefficients. The coefficient \( a \) should not be zero. This type of equation is called 'quadratic' because the variable \( x \) is raised to the power of two. A quadratic equation can have up to two solutions, which can be found using various methods, such as factoring, completing the square, or the quadratic formula.

The quadratic formula is particularly useful for solving quadratic equations. It is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

- The "\( \pm \)" symbol indicates that there can be two potential solutions.
- The term under the square root, \( b^2 - 4ac \), is known as the discriminant. It tells us about the nature of the roots of the quadratic equation.

If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one real root (or a repeated root), and if it is negative, the equation has no real roots, but two complex ones.

System of Equations

A system of equations consists of multiple equations, typically involving more than one variable, used together to determine unknown values. These systems can take different forms:

- **Linear systems:** Involving equations of first degree. They generally produce straight lines when graphed.
- **Nonlinear systems:** Include at least one equation that is not linear, like our problem given with quadratic equations, leading to curves when graphed.

To solve a system of equations, one can employ a variety of strategies such as substitution, elimination, or graphical methods. The core goal is to find a common solution (or solutions) that satisfies all equations simultaneously. In the context of the problem given, we were working with a nonlinear system, since one equation was quadratic.

Real-world applications often involve systems of equations, making them crucial in fields like economics, engineering, and physics. Depending on the equations' nature, the system could have a single solution, multiple solutions, or potentially no solution at all.

Substitution Method

The substitution method is a frequently used technique for solving systems of equations. Here's a step-by-step explanation of how the method works, specifically in the context of our provided system of equations:- **Identify and Isolate:** Choose an equation where one variable can be easily isolated. For example, in the system \( x = y + 2 \), \( x \) is already isolated.- **Substitute:** Once a variable is isolated, replace that variable in the other equation(s) with the expression you found. This step generates a new equation with only one variable. From our exercise, by substituting \( x = y + 2 \) into the first equation, we form a single-variable equation in terms of \( y \).- **Solve for One Variable:** Use algebraic methods or formulas like the quadratic formula to solve the new equation for one variable.- **Back-Substitute:** Once the value(s) for one variable are found, substitute these back into the isolated equation to find the value(s) of the other variable.The substitution method is powerful when there is a clear opportunity to isolate one variable, making it straightforward to reduce the system to a single-variable problem that can be solved easily. Remember, practice with this method will help smooth out the process, making it a reliable tool in solving systems of equations.