Which inequality has solution set (-∞, ∞)? A. (x-3)^2≥0 B. (5x-6)^2≤0 C. (6x+4)^2>0 D. (8x+7)^2<0

Verified step by step guidance

1. Step 1: Understand the problem. We need to determine which inequality has a solution set that includes all real numbers, represented by (-∞, ∞).
2. Step 2: Analyze each option. Start with option A: $(x-3{)}^2\ge 0$. Since a square of any real number is always non-negative, this inequality is true for all real numbers.
3. Step 3: Consider option B: $(5x-6{)}^2\le 0$. A square of a real number is zero or positive, so it can only be zero. This means $5x-6=0$, which is a single solution, not all real numbers.
4. Step 4: Evaluate option C: $(6x+4{)}^2>0$. A square is positive for all real numbers except when it is zero. This means $6x+4\ne 0$, which excludes one value, not all real numbers.
5. Step 5: Examine option D: $(8x+7{)}^2<0$. A square of a real number cannot be negative, so there are no real solutions.

Key Concepts

Inequalities

Inequalities are mathematical statements that compare two expressions, indicating that one is greater than, less than, or equal to the other. They can be strict (using < or >) or non-strict (using ≤ or ≥). Understanding how to manipulate and solve inequalities is crucial for determining solution sets.

Quadratic Expressions

Quadratic expressions are polynomials of degree two, typically in the form ax^2 + bx + c. The sign of a quadratic expression can vary depending on the values of x, and its roots (where it equals zero) play a key role in determining the intervals where the expression is positive or negative.

Solution Sets

A solution set is the collection of all values that satisfy a given inequality or equation. For example, the solution set (-∞, ∞) indicates that all real numbers are solutions, which occurs when the inequality is always true, such as when a squared term is greater than or equal to zero.