Problem 29 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

No real solution exists for this nonlinear system.

Step 1: Understand the System

We have two equations: \(-x^2 + y = 2\) and \(2y = -x\). Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve for y in the Second Equation

The second equation is \(2y = -x\). We can express \(y\) in terms of \(x\) by dividing each side by 2: \(y = -\frac{x}{2}\).

Step 3: Substitute y in the First Equation

Now that we have \(y = -\frac{x}{2}\), substitute this expression into the first equation: \(-x^2 + (-\frac{x}{2}) = 2\). Simplify to obtain \(-x^2 - \frac{x}{2} = 2\).

Step 4: Eliminate the Fraction

Multiply every term by 2 to eliminate the fraction: \(-2x^2 - x = 4\).

Step 5: Rearrange into a Quadratic Equation

Bring all terms to one side: \(-2x^2 - x - 4 = 0\). Rearrange into standard form: \(2x^2 + x + 4 = 0\).

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = 4\). Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot 4 = 1 - 32 = -31\). Since the discriminant is negative, there are no real solutions for \(x\).

Step 7: Interpret the Solution

Since there are no real solutions for \(x\), this implies that there are no real values for \(x\) and \(y\) that satisfy both equations at the same time.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest degree is two. A standard quadratic equation is represented as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The graph of a quadratic equation is a parabola.
Understanding quadratic equations is essential because they frequently appear in various aspects in algebra and geometry.

• The term \(ax^2\) represents the quadratic part and determines the parabola's width and direction (upwards or downwards).
• The term \(bx\) represents the linear part, affecting the parabola's slope and orientation.
• The constant term \(c\) moves the parabola up or down the y-axis.

In our problem, we manipulated the system to form the quadratic equation \(2x^2 + x + 4 = 0\) through substitution, which we'll discuss further.

Substitution Method

The substitution method is a useful technique for solving systems of equations, particularly when one equation is linear. The method involves expressing one variable in terms of another and substituting this into the other equation.
This reduces the system to a single equation with only one variable.

• In our exercise, the second equation was linear (\(2y = -x\)), making it easy to express \(y\) in terms of \(x\).
• We then substituted \(y = -\frac{x}{2}\) into the quadratic equation, resulting in \(-x^2 - \frac{x}{2} = 2\).

This step eliminated one variable and allowed us to solve for \(x\) by converting it into a quadratic equation. Always double-check your substitution to make sure it's correct, as errors here can lead to incorrect results.

Discriminant in Quadratic Formula

The discriminant is a component of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is calculated as \(b^2 - 4ac\), where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
Here's how the discriminant works:

• If the discriminant is positive, the equation has two distinct real roots.
• If it's zero, the equation has exactly one real root (or a repeated root).
• If it's negative, like in our problem \(-31\), there are no real roots, only complex ones.

The discriminant told us there were no real solutions in our exercise, guiding us to conclude the system of equations has no real solution.

No Real Solutions

When a quadratic equation has no real solutions, it means that the associated parabola does not intersect the x-axis. This can occur when solving a system of nonlinear equations, indicating that the equations do not share a common real solution.
When we ended up with a negative discriminant, it informed us that there are no real values for \(x\) and correspondingly no real \(y\) that satisfy both the given equations.

• Interpret this algebraically: It implies that, graphically, the solutions lie outside the realm of real numbers.
• Practically, no real intersection point exists for the graphed equations.

Understanding this concept is crucial for identifying when real solutions are not possible and being able to adjust your solution approach accordingly.