Problem 49 - Precalculus : Building Concepts and Connections

Question

Explain why \((x+1)^{2} \leq 0\) has a solution, whereas \((x+1)^{2}<0\) has no real-valued solution.

Step-by-Step Solution

Answer

The inequality \((x+1)^{2}<0\) has no real-valued solutions because a square of a number can never be less than zero while \((x+1)^{2} \leq 0\) has a solution at x=-1 as squares values can indeed be zero.

Step 1: Understand the Inequalities

The first inequality says \((x+1)^{2} \leq 0\), which means the value of \((x+1)^2\) can be either 0 or less than 0. The second inequality \((x+1)^{2}<0\) states that \((x+1)^2\) must be less than 0.

Step 2: Analyze the Properties of Square Numbers

Recall that the square of a number, whether it is positive or negative, is always positive or zero. This is due to the property of square numbers that \((–a)^2 = a^2\), so no squared number can be less than zero.

Step 3: Apply Square Number Property to the Inequalities

Considering the property of square numbers, where no square can be negative, you can see that \((x+1)^{2}<0\) has no solution because a square can never be negative. In contrast, \((x+1)^{2} \leq 0\) does have a solution, which is x=-1, because a squared number can equal zero.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that take the form \(ax^2 + bx + c = 0\). The main characteristic of these equations is that they involve the second power of the variable \(x\). To solve them, we often use methods such as factoring, completing the square, or the quadratic formula. The solutions for \(x\) in these equations may be real numbers, complex numbers, or sometimes repeated values. Quadratic equations are foundational in algebra because they frequently arise in various problem-solving scenarios, including physics, engineering, and everyday calculations.

In the exercise given, although it isn't a standard quadratic equation, it directly relates to such equations since it requires solving an expression where \((x+1)^2\) is involved. Understanding how to manipulate squares and inequalities is crucial for dealing with problems of this nature.

Real Number Solutions

Real number solutions are the values of \(x\) that satisfy an equation within the set of real numbers, a comprehensive set that includes all rational and irrational numbers. In the context of quadratic equations, whether the solutions are real depends largely on the discriminant (\(b^2 - 4ac\)).

When handling inequalities like \( (x+1)^2 \leq 0\) and \( (x+1)^2 < 0\), it is essential to determine if such solutions fall into the realm of real numbers. For the first inequality \( (x+1)^2 \leq 0\), the solution \(x = -1\) exists as a point on the real number line where the expression equals zero. For the second inequality, however, the expression \( (x+1)^2 < 0\) cannot have real solutions because, as will be explained, no real number squared gives a negative result.

Properties of Square Numbers

The properties of square numbers form a critical basis for understanding why an inequality like \( (x+1)^2 < 0\) has no real solutions. Squares of numbers, whether positive or negative, yield positive results or zero, represented by the equation \( y^2 \geq 0\) for any real number \(y\). This property holds because multiplying two negative numbers or two positive numbers will produce a positive outcome: \( (-a)^2 = a^2\).

Given this key property, when analyzing the exercise, the inequality \( (x+1)^2 \leq 0\) has a solution only at \(x = -1\), where the square is exactly zero. However, \( (x+1)^2 < 0\) has no solution among real numbers because a square cannot be negative. This logical impasse reinforces this property of squares, establishing the limits of real solutions constraints given by squaring.