Math Problem Statement

What is the solution set of the quadratic inequality 4(x + 2)^2 ≤ 0?

Solution

Let's solve the quadratic inequality 4(x+2)204(x + 2)^2 \leq 0.

Step-by-step solution:

  1. Simplify the inequality:

    The expression can be rewritten as:

    4(x+2)204(x + 2)^2 \leq 0

    Divide both sides by 4:

    (x+2)20(x + 2)^2 \leq 0

  2. Interpret the inequality:

    • The square of a real number is always non-negative, meaning (x+2)20(x + 2)^2 \geq 0 for all xRx \in \mathbb{R}.
    • The only way (x+2)2=0(x + 2)^2 = 0 is if x+2=0x + 2 = 0, which gives x=2x = -2.
  3. Conclusion:

    The inequality (x+2)20(x + 2)^2 \leq 0 holds only when x=2x = -2. For any other value of xx, (x+2)2(x + 2)^2 will be positive, and thus the inequality would not hold.

Thus, the solution set is {xx=2}\{ x \mid x = -2 \}.

Answer:

The correct answer is the first option: {xx=2}\{ x \mid x = -2 \}.


Would you like more details or have any questions? Here are 5 related questions to expand on this:

  1. How would the solution change if the inequality were 4(x+2)2<04(x + 2)^2 < 0?
  2. What are the general steps to solve quadratic inequalities?
  3. How would you solve 4(x+2)204(x + 2)^2 \geq 0?
  4. What is the difference between solving an inequality and an equation?
  5. Can the inequality 4(x+2)204(x + 2)^2 \leq 0 have more than one solution?

Tip: Always remember that the square of any real number is non-negative. This fact is key in solving many quadratic inequalities.

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Math Problem Analysis

Mathematical Concepts

Quadratic Inequalities
Algebra

Formulas

(x + 2)^2 = 0

Theorems

Non-negativity of squares

Suitable Grade Level

Grades 9-12