Problem 62 - College Algebra with Modeling and Visualization

Question

If possible, solve the nonlinear system of equations. $$ \begin{array}{rr} x^{2}+y^{2}= & 4 \\ 2 x^{2}+y & =-3 \end{array} $$

Step-by-Step Solution

Answer

The system admits real solutions which are the intersection points computed from the circle and parabola.

Step 1: Analyze the System

We have two equations: \(x^2 + y^2 = 4\) which represents a circle with radius 2 centered at the origin, and \(2x^2 + y = -3\), which is an equation that involves a parabola. We will solve these equations simultaneously to find the points of intersection.

Step 2: Express One Variable in Terms of the Other

From the second equation, isolate \(y\): \(y = -3 - 2x^2\). Substitute this expression for \(y\) into the first equation to eliminate \(y\).

Step 3: Substitute and Simplify

Substitute \(y = -3 - 2x^2\) into \(x^2 + y^2 = 4\):\[ x^2 + (-3 - 2x^2)^2 = 4 \] Expand and simplify the equation:\[ x^2 + (9 + 12x^2 + 4x^4) = 4 \]\[ 4x^4 + 13x^2 + 9 = 4 \]\[ 4x^4 + 13x^2 + 5 = 0 \].

Step 4: Solve the Polynomial Equation

Let \(u = x^2\), then the equation becomes \(4u^2 + 13u + 5 = 0\). Use the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 4, b = 13, c = 5\).Calculate the discriminant \(b^2 - 4ac = 169 - 80 = 89\), which is positive, indicating two real roots. Compute:\[ u = \frac{-13 \pm \sqrt{89}}{8} \].

Step 5: Compute the Values of \(u=x^2\) and \(x\)

Calculate the roots for \(u\):\[ u_1 = \frac{-13 + \sqrt{89}}{8}, \quad u_2 = \frac{-13 - \sqrt{89}}{8} \].Since \(u\) represents \(x^2\), it must be non-negative. Determine the valid \(u\) that is \(u_1\) since \(u_2\) is negative.

Step 6: Determine Corresponding \(y\) Values

Find \(x\) from \(x^2 = u_1\):Compute \(x\) as \(\pm \sqrt{u_1}\). Then substitute back into \(y = -3 - 2x^2\) to find \(y\) values for each \(x\).

Step 7: Solutions Summary

Evaluate both possibilities for \(x\) and compute respective \(y\) values. Conclude with the \((x, y)\) pairs that satisfy both equations.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). It's particularly useful when factoring is complex or difficult. The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. The part under the square root, \(b^2 - 4ac\), is called the discriminant. The discriminant determines the nature of the roots:

• If it's positive, there are two distinct real solutions.
• If it's zero, there is exactly one real solution.
• If it's negative, the equation has two complex solutions.

In our exercise, after substituting variables and simplifying the non-linear system, we derived a quadratic equation in terms of \(u\) (where \(u = x^2\)). Using the quadratic formula allowed us to solve for \(u\) efficiently, identifying the potential values for \(x^2\). This step was crucial because it opened the door to finding possible real \(x\) values needed for solving the intersection problem.

System of Equations

A system of equations consists of two or more equations with common variables. The main goal is to find values for these variables that satisfy all given equations simultaneously. There are various methods to solve these systems:

• Substitution: Solve one equation for one variable and substitute this expression into the other equation(s).
• Elimination: Combine equations to eliminate one variable, simplifying the system.
• Graphical Method: Plotting equations to identify points of intersection.

In our exercise, we combined substitution and algebraic manipulation. We isolated \(y\) in the second equation and substituted it into the first. This allowed us to convert a potentially complicated nonlinear system into a single polynomial equation in terms of \(x\). Solving this equation provided a path forward in addressing the intersection of a circle and a parabola.

Circle and Parabola Intersection

In geometry, finding the intersection of a circle and a parabola involves solving their equations simultaneously. The circle's equation is typically of the form \(x^2 + y^2 = r^2\), where \(r\) is the radius. The parabola's equation can vary but typically involves terms like \(y = ax^2 + bx + c\).For intersections, the solutions are points that lie on both curves. Each point of intersection represents a solution to the system of equations derived from both shapes. Here's how the process unfolds:

• Isolate one variable in one equation (substitution).
• Replace that variable in the other equation.
• Solve the resulting equation to find the potential \(x\) (or \(y\)) values.

In our case, after substituting and simplifying, we reached a polynomial equation. Solving it using the quadratic formula gave potential \(x\) values. Substituting these back into one of the original equations yielded corresponding \(y\) values, allowing us to determine the precise points of intersection between the circle \(x^2 + y^2 = 4\) and the parabola \(2x^2 + y = -3\). The solution set of points represents where both curves intersect on a plane.