Problem 29 - College Algebra

Question

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

Step-by-Step Solution

Answer

The system has no real solution as the quadratic discriminant is negative.

Step 1: Identify the system of equations

The system of equations we need to solve is given by the two equations: 1. \(-x^2 + y = 2\)2. \(2y = -x\)

Step 2: Solve the linear equation for one variable

Taking the second equation, \(2y = -x\), let's solve for \(y\):\[ y = -\frac{x}{2} \]

Step 3: Substitute into the nonlinear equation

Now, substitute \(y = -\frac{x}{2}\) into the first equation:\[ -x^2 + \left(-\frac{x}{2}\right) = 2 \]

Step 4: Simplify and solve the quadratic equation

Simplify the equation:\[ -x^2 - \frac{x}{2} = 2 \]Multiply every term by \(-1\):\[ x^2 + \frac{x}{2} = -2 \]Multiply through by 2 to eliminate the fraction:\[ 2x^2 + x = -4 \]Rearrange to form a standard quadratic equation:\[ 2x^2 + x + 4 = 0 \]

Step 5: Solve the quadratic equation

Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), \(c = 4\).Calculate the discriminant:\[ b^2 - 4ac = 1^2 - 4(2)(4) = 1 - 32 = -31 \]Since the discriminant is negative, the quadratic equation has no real roots.

Step 6: Conclude whether there is a solution

Since there are no real solutions for \(x\) from the quadratic equation, it implies that there are no real solutions for the system of equations given.

Key Concepts

Nonlinear Systems

A system of equations is said to be nonlinear if at least one of the equations is not a straight line when graphed. This could mean it involves variables raised to powers other than one, like squares in quadratic equations. In our example, the system is comprised of one nonlinear and one linear equation:

• The nonlinear equation is \(-x^2 + y = 2\), where \(x\) is raised to the power of 2, indicating that it is quadratic in nature.
• The linear equation is \(2y = -x\), a straightforward line where neither variable is squared or otherwise complex.

Solving nonlinear systems often involves techniques like substitution or elimination. In these cases, solving one equation for a variable and substituting it into the other equation helps simplify the process. In many math problems, especially those involving curves like parabolas, this method makes it easier to see potential solutions. By isolating one variable, computations become more manageable and revealing any potential intersections or solutions comes more straightforward.

Quadratic Equations

A quadratic equation typically appears in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Quadratic equations graph as parabolic curves. They can open upwards or downwards depending on the sign of \(a\). In the given task, the equation in question transforms into:

• \(x^2 + \frac{x}{2} = -2\)
• Cleaning up to \(2x^2 + x + 4 = 0\) through simplification.

Typically, you solve these equations using methods like factoring, completing the square, or applying the quadratic formula. These processes reveal the roots or x-values by setting the equation to zero. Solving this type of equation exposes the x-values where the parabola crosses the x-axis. Roots could be real, complex or repeated, based on the discriminant value, giving an idea of intersection points when combined with linear systems.

Discriminant

The discriminant is a component of the quadratic formula, expressed as \(b^2 - 4ac\). It plays a crucial role in determining the nature of the solutions of a quadratic equation:

• If the discriminant is positive, the equation has two distinct real solutions.
• If it equals zero, there is exactly one real solution, called a repeated root.
• A negative discriminant indicates no real solutions; the roots are complex or imaginary.

In our problem, the discriminant calculation \(1^2 - 4(2)(4) = -31\) results in a negative value. Thus, the quadratic equation has no real solutions; instead, it suggests complex roots. This directly influences our original system of equations, signifying there is no point of intersection between the parabola and the line in the real plane, confirming no real solution exists for both equations together.