Problem 21 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{r} x^{2}+y^{2}=4 \\ -2 x^{2}+3 y=6 \end{array}$$

Step-by-Step Solution

Answer

The approximate points of intersection between the given circle and line are: \((0.99, 1.42)\) and \((0.99, -1.42)\).

Step 1: Solve the first equation for one variable

First, we will solve the first equation (circle) for one variable (either x or y). In this case, we will solve for y. From the first equation: \(x^2 + y^2 = 4\) Solving for y, we get: \(y^2 = 4 - x^2\) \(y = \pm\sqrt{4 - x^2}\)

Step 2: Substitute the expression for y in the second equation

Now, substitute the expression for y from step 1 into the second equation: \(-2x^2 + 3(\pm\sqrt{4 - x^2}) = 6\) Now, we have two cases to consider: Case 1 (positive value of y): \( -2x^2 + 3\sqrt{4 - x^2} = 6\) Case 2 (negative value of y): \( -2x^2 -3\sqrt{4 - x^2} = 6\)

Step 3: Solve each case for x

Case 1: \(-2x^2 + 3\sqrt{4 - x^2} = 6\) Add \(2x^2\) to both sides: \(3\sqrt{4 - x^2} = 6 + 2x^2\) Divide both sides by 3: \(\sqrt{4 - x^2} = 2 + \frac{2}{3}x^2\) Square both sides: \(4 - x^2 = 4 + \frac{8}{3}x^2 + 4x^4\) Move all terms to the right: \(x^2 + 4x^4 + \frac{8}{3}x^2 - 4 = 0\) This is a quartic equation which is difficult to solve by standard methods. However, we can make the observation that \(x^2 \leq 4\) since the given circle has a radius of 2. Therefore, the quartic term is small compared to the other terms (specifically, it's \(1/4\) of the size of quadratic term), so we can make the approximation: \(x^2 + \frac{8}{3}x^2 - 4 \approx 0\) Now we have a quadratic equation with the standard form \(ax^2+bx+c = 0\), which can be solved using Quadratic Formula: \(x^2 \left(1+\frac{8}{3}\right) - 4 = 0\) \(x^2 \left(\frac{11}{3}\right) - 4 = 0\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get: \(x = \frac{0 \pm \sqrt{0 - 4\left(\frac{11}{3}\right)(-4)}}{2\left(\frac{11}{3}\right)}\) \(x = \frac{\pm\sqrt{\frac{176}{3}}}{\frac{22}{3}}\) \(x \approx \pm 0.99\) Now we go back to the second case. Case 2: \(-2x^2 - 3\sqrt{4 - x^2} = 6 \) Using the same steps as in case 1, we are left with the equation \(x^2 + 4x^4 -\frac{8}{3}x^2 + 4 = 0\) Applying the same approximation as in the case 1, we have: \(x^2 - \frac{8}{3}x^2 + 4 \approx 0\) \(x^2\left(1-\frac{8}{3}\right) + 4 = 0\) In this case, the coefficient of the quadratic term is negative, there are no real solutions for x. Thus, this case has no solutions.

Step 4: Find the y-coordinates

From the approximated values of x, we can now find the corresponding y-coordinates: For \(x \approx 0.99\) and \(y = \pm\sqrt{4 - x^2}\), \(y \approx \pm\sqrt{4 - (0.99)^2} \approx \pm 1.42\)

Step 5: List the solutions

So, the approximate points of intersection between the given circle and line are: \((0.99, 1.42)\) and \((0.99, -1.42)\)

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations form a parabola when graphed. The solutions to these equations correspond to the points where the parabola intersects the x-axis. To solve quadratic equations, we typically use methods like factoring, completing the square, or the quadratic formula.

In the given problem, we encounter the equation \(x^2 + y^2 = 4\), which represents a circle. This is not typical for quadratic equations but highlights their versatility in representing various curves. The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), aids in finding the solutions. It’s essential to recognize the structure of quadratic equations to solve them correctly.

Solving Systems Algebraically

When solving a system of equations algebraically, it means finding values for variables that satisfy all equations in the system simultaneously. We use substitution or elimination methods to do this.

• **Substitution**: Solve one of the equations for one variable and substitute this expression into the other equation(s).
• **Elimination**: Add or subtract equations to eliminate one of the variables, making it easier to solve for the remaining variable(s).

In this exercise, substitution was used. First, we solved one of the equations for \(y\) to express it in terms of \(x\), then substituted this expression in the other equation. This transformation simplifies the problem to a single variable equation, allowing for easier manipulation and solution. This method ensures that all variables are considered in relation to each other, essential for finding all possible solutions.

Mathematical Approximation

Mathematical approximation is used when obtaining exact solutions is complicated or impossible. Approximations simplify equations so that they can be solved more easily or used in practical applications where precision might not be critical.

For example, when solving the equation \(-2x^2 + 3\sqrt{4 - x^2} = 6\), complications arise with the quartic terms, making an exact solution difficult. By recognizing that the higher-order terms are relatively small, they are approximated or ignored, simplifying the equation: \(x^2 + \frac{8}{3}x^2 - 4 \approx 0\). This reduces the complexity and allows the use of the quadratic formula, providing an approximate solution. Approximations are crucial in practical scenarios where complex equations arise, making solutions feasible to determine behaviors or events realistically.