Problem 52 - College Algebra

Question

Find the solutions to the nonlinear equations with two variables. $$ \begin{aligned} x^{2}+4 x y-2 y^{2}-6 &=0 \\ x &=y+2 \end{aligned} $$

Step-by-Step Solution

Answer

\( x = \frac{\sqrt{42}}{3}, y = -2 + \frac{\sqrt{42}}{3} \) and \( x = -\frac{\sqrt{42}}{3}, y = -2 - \frac{\sqrt{42}}{3} \).

Step 1: Substitute Linear Equation into Nonlinear Equation

Given the linear equation \( x = y + 2 \), substitute this expression for \( x \) into the nonlinear equation \( x^2 + 4xy - 2y^2 - 6 = 0 \). This leads to \( (y+2)^2 + 4(y+2)y - 2y^2 - 6 = 0 \).

Step 2: Expand and Simplify

Expand the expression from Step 1: \( (y+2)^2 = y^2 + 4y + 4 \) and \( 4(y+2)y = 4(y^2 + 2y) = 4y^2 + 8y \). Substitute these into the equation to get \( y^2 + 4y + 4 + 4y^2 + 8y -2y^2 - 6 = 0 \). Combine like terms to 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, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = 12 \), and \( c = -2 \). Calculate \( y = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 3 \cdot (-2)}}{6} \). Simplify to find the possible values of \( y \).

Step 4: Calculate the Discriminant and Simplify

Calculate the discriminant: \( 12^2 - 4 \times 3 \times (-2) = 144 + 24 = 168 \). Now, \( y = \frac{-12 \pm \sqrt{168}}{6} = \frac{-12 \pm 2\sqrt{42}}{6} \). Further simplify to \( y = -2 \pm \frac{sqrt{42}}{3} \).

Step 5: Solve for x Using Values of y

Substituting these \( y \) values back into \( x = y + 2 \), for \( y = -2 + \frac{\sqrt{42}}{3} \), \( x = 0 + \frac{\sqrt{42}}{3} \), and for \( y = -2 - \frac{\sqrt{42}}{3} \), \( x = 0 - \frac{\sqrt{42}}{3} \).

Step 6: Verify the Solutions

Verify solutions by plugging them back into the original nonlinear equation. This step involves substituting \( (x, y) = \left( \frac{ \sqrt{42} }{3}, -2 + \frac{ \sqrt{42} }{3} \right) \) and \( (x, y) = \left( -\frac{ \sqrt{42} }{3}, -2 - \frac{ \sqrt{42} }{3} \right) \) into the equation \( x^2 + 4xy - 2y^2 - 6 = 0 \) to confirm both satisfy the equation.

Key Concepts

Quadratic Equation

Quadratic equations are a special type of algebraic equation characterized by a degree of two. This means the highest power of the variable is squared (i.e., raised to the power of two). The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] where:

• \( a, b, \text{ and } c \) are constants, with \( a eq 0 \).
• \( x \) represents the unknown variable.

In the context of our problem, after substitution and simplification (which we will discuss next), the equation \( 3y^2 + 12y - 2 = 0 \) is a quadratic equation in terms of \( y \). Solving quadratic equations often involves methods such as factoring, completing the square, or using the quadratic formula. In this case, we used the quadratic formula to find the values of \( y \).

Substitution Method

The substitution method is a powerful tool for solving systems of equations, particularly when working with nonlinear equations. The idea is to express one variable in terms of another, using one of the given equations, and then substitute this expression into the remaining equations. This reduces the number of variables, typically simplifying the problem.In our exercise, we began with the linear equation \( x = y + 2 \). By substituting \( y + 2 \) for \( x \) in the nonlinear equation \( x^2 + 4xy - 2y^2 - 6 = 0 \), the equation was transformed entirely in terms of \( y \). This reduction makes the solution more manageable by allowing us to focus on solving a single-variable quadratic equation.

Discriminant

The discriminant is a crucial concept in the context of solving quadratic equations, particularly when using the quadratic formula. The discriminant is part of the quadratic formula and is expressed as:\[ b^2 - 4ac \]It reveals important information 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.
• If the discriminant is negative, the roots are complex (not real).

In our problem, the calculation of the discriminant, \( 12^2 - 4 \times 3 \times (-2) = 168 \), indicates that the equation has two distinct real roots. This positive value signified that the solutions would be real numbers, specifically prompting us to find the square root of the discriminant in the next step.

Verify Solutions

Verifying the solutions after solving an equation ensures the accuracy of the obtained results, confirming they satisfy the original equations. This step involves back-substitution, which means plugging the calculated values of \( x \) and \( y \) back into the original equation.In our exercise, the solutions found were:

• \( (x, y) = \left( \frac{ \sqrt{42} }{3}, -2 + \frac{ \sqrt{42} }{3} \right) \)
• \( (x, y) = \left( -\frac{ \sqrt{42} }{3}, -2 - \frac{ \sqrt{42} }{3} \right) \)

We verified these solutions by substituting them back into the nonlinear equation \( x^2 + 4xy - 2y^2 - 6 = 0 \) and confirming that both pairs satisfy the equation. This final check ensures that there are no errors in computations or logic throughout the problem-solving process.