Problem 2 - Algebra 2 and Trigonometry

Question

Explain why the equations \(y=x^{2}+2\) and \(y=-2\) have no common solution in the set of real numbers.

Step-by-Step Solution

Answer

The equations have no common real solutions because the quadratic \(x^2 = -4\) is impossible with real numbers.

Step 1: Set Equations Equal to Find Intersection Points

To determine if the two equations have common solutions, set them equal to each other: \[ x^2 + 2 = -2 \].

Step 2: Simplify the Equation

Subtract 2 from both sides to isolate the quadratic expression:\[ x^2 + 2 - 2 = -2 - 2 \]which simplifies to\[ x^2 = -4 \].

Step 3: Analyze the Quadratic Equation

Since \(x^2 = -4\), we need to find the real number \(x\) such that \(x^2\) is a negative number. Recall that any real number squared (\(x^2\)) is always non-negative, i.e., \(x^2 \geq 0\). Thus, \(x^2 = -4\) has no real solutions.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are shaped like parabolas when graphed. Quadratic equations can have different types of solutions depending on the discriminant, \(b^2 - 4ac\).

- If the discriminant is positive, the quadratic equation has two distinct real solutions.
- If the discriminant is zero, there is exactly one real solution, also known as a repeated solution.
- If the discriminant is negative, there are no real solutions, although there might be two complex solutions.

Understanding how to identify these solutions helps in determining the nature of the equations, like whether their graphs intersect or not.

Intersection Points

Intersection points occur when two graphs meet or cross each other. To find these points for two equations, we set the equations equal to each other and solve for the variable.

In the case of the given equations \(y = x^2 + 2\) and \(y = -2\), setting them equal gives us \(x^2 + 2 = -2\). Simplifying this equation leads to \(x^2 = -4\).

Since no real number squared results in a negative number, these two equations do not intersect at any real point. Therefore, there are no intersection points between their graphs in the plane of real numbers.

Real Numbers

Real numbers include all the numbers on the number line such as positive numbers, negative numbers, and zero. They can be whole numbers, fractions, and irrationals like \(\sqrt{2}\) and \(\pi\). But a key feature of real numbers when it comes to solving quadratic equations is that squaring any real number never results in a negative value.

This is crucial for understanding why the equation \(x^2 = -4\) lacks solutions in the real number set. Since squaring any real number results in a positive or zero, \(x^2 = -4\) is impossible to satisfy with real numbers, leading to no real solution.

Negative Solutions

Negative solutions are not possible when a variable squared equals a number less than zero in the set of real numbers. For any real number \(x\), squaring \(x\) means multiplying it by itself, which yields either a positive number or zero. Negative solutions arise in complex numbers involving imaginary units, where the imaginary unit \(i\) satisfies \(i^2 = -1\).

In the context of our problem, when faced with \(x^2 = -4\), we see why it cannot be solved with real numbers. The only numbers that can satisfy such an equation are complex numbers such as \(x = 2i\) or \(x = -2i\). However, within the real number set required by the problem, negative solutions are non-existent, reaffirming the lack of intersection between the given functions.