Problem 51 - Precalculus: Functions and Graphs

Question

Solve the system of equations graphically and algebraically. Compare your answers. $$x^{2}+y^{2}=4 ; \quad x+y=1$$

Step-by-Step Solution

Answer

The solutions are \((\frac{1 + \sqrt{7}}{2}, 1 - \frac{1 + \sqrt{7}}{2})\) and \((\frac{1 - \sqrt{7}}{2}, 1 - \frac{1 - \sqrt{7}}{2})\).

Step 1: Understanding the System of Equations

The system consists of two equations: the first is a circle, \(x^2 + y^2 = 4\), and the second is a line, \(x + y = 1\). We need to find their points of intersection, i.e., the solutions to the system.

Step 2: Graphical Solution

Graph the circle \(x^2 + y^2 = 4\), which is centered at the origin \((0,0)\) with radius 2. Then graph the line \(x + y = 1\), which intersects the y-axis at (0,1) and the x-axis at (1,0). The points of intersection of these graphs are the solutions to the system.

Step 3: Substitute to Solve Algebraically

Since \(x + y = 1\), express \(y\) in terms of \(x\): \(y = 1 - x\). Substitute \(y = 1 - x\) into \(x^2 + y^2 = 4\):\[x^2 + (1-x)^2 = 4\].

Step 4: Solving the Quadratic Equation

Expand and simplify the equation:\[x^2 + (1-x)^2 = 4\] which simplifies to \[x^2 + 1 - 2x + x^2 = 4\]. Combine like terms to get \[2x^2 - 2x + 1 = 4\]. Simplify to \[2x^2 - 2x - 3 = 0\].

Step 5: Factoring the Quadratic

Divide the entire equation by 2 to simplify: \[x^2 - x - \frac{3}{2} = 0\]. Since this is hard to factor directly, it seems more practical to directly find numeric solutions using either factoring or the quadratic formula.

Step 6: Quadratic Formula

Solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -1\), and \(c = -\frac{3}{2}\). Substitute these values into the formula to find \(x\).

Step 7: Calculate the Roots

Calculate: \[x = \frac{1 \pm \sqrt{1 + 6}}{2}\]. This simplifies to \[x = \frac{1 \pm \sqrt{7}}{2}\], giving two solutions for \(x\): \(x = \frac{1 + \sqrt{7}}{2}\) and \(x = \frac{1 - \sqrt{7}}{2}\).

Step 8: Find 'y' Values

Using \(y = 1 - x\), find the corresponding \(y\) values for each \(x\). For \(x = \frac{1 + \sqrt{7}}{2}\), \(y = 1 - \frac{1 + \sqrt{7}}{2}\). For \(x = \frac{1 - \sqrt{7}}{2}\), \(y = 1 - \frac{1 - \sqrt{7}}{2}\).

Step 9: Compare Algebraic and Graphical Solutions

Graphing gives two points of intersections, \((\frac{1 + \sqrt{7}}{2}, 1 - \frac{1 + \sqrt{7}}{2})\) and \((\frac{1 - \sqrt{7}}{2}, 1 - \frac{1 - \sqrt{7}}{2})\), same as the solutions calculated algebraically.

Key Concepts

Graphical Solution

A graphical solution to a system of equations involves plotting the equations on a graph to find their intersection points. These intersections represent the solutions. For this system, we have a circle and a line.

• **Circle:** The equation is given by \(x^2 + y^2 = 4\). This describes a circle centered at the origin \((0,0)\) with a radius of 2, meaning the circle reaches out to 2 units in all directions from the center.
• **Line:** The equation is \(x + y = 1\). This is a straight line that crosses the y-axis at (0,1) and the x-axis at (1,0). These are its intercepts.

By drawing both the circle and the line on the same graph, the solution to the system is found at the intersection points of the line and the circle. In this instance, the graphical solution revealed two points of intersection.

Circle Equation

Circle equations in algebra follow the standard form: \(x^2 + y^2 = r^2\), where \(r\) is the radius. Here, we analyze the circle defined by this equation:

• **Center:** The center of any circle given by \(x^2 + y^2 = r^2\) is at the origin (0,0), because there are no shifts indicated by constants accompanying x and y.
• **Radius:** The radius is given directly by the square root of the constant on the right side. For example, \(x^2 + y^2 = 4\) means the radius \(r = \sqrt{4} = 2\).

Knowing the center and the radius allows for accurate graphic representation, forming the basis for a graphical solution in many geometric contexts.

Line Equation

A line equation typically appears in the form of \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. However, in our system's form, \(x + y = 1\):

• **Slope-Intercept Form:** You can rearrange it to \(y = -x + 1\). Here, \(m = -1\), indicating a descending line.
• **Intercepts:** The intercepts of the line, found initially by substitution, are (0,1) and (1,0). These are where the line meets the y and x axes, respectively.

These concepts allow for the visualization of the line when solving graphically, and the slope tells us the angle of descent across the graph.

Quadratic Equation

A quadratic equation is generally written as \(ax^2 + bx + c = 0\). To solve for \(x\), we might use factoring, completing the square, or the quadratic formula. In this exercise, the solution involves using the quadratic formula:

• **Quadratic Formula:** \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This equation provides solutions for \(x\) by evaluating the expression under the square root, known as the discriminant.
• **Substitution:** In our case, to find the intersection points, \(x\) is expressed through \(x^2 + (1-x)^2 = 4\), which simplifies to a quadratic with \(a = 1\), \(b = -1\), and \(c = -\frac{3}{2}\).

Solving this gives us the points where the circle and line intersect. Quadratic equations provide one powerful algebraic method to derive solutions that can be cross-verified with graphical approaches.