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What is the solution set of the quadratic inequality [tex]$4(x+2)^2 \leq 0$[/tex]?

A. [tex]\{x \mid x = -2\}[/tex]

B. [tex]\{x \mid x \in \mathbb{R}\}[/tex]

C. [tex]\{x \mid x = 2\}[/tex]

D. [tex]\varnothing[/tex]


Sagot :

Certainly! Let's solve the quadratic inequality [tex]\(4(x+2)^2 \leq 0\)[/tex] step by step.

1. Understand the inequality: You are given the quadratic inequality [tex]\(4(x + 2)^2 \leq 0\)[/tex].

2. Solve the corresponding equation: To solve this inequality, we first solve the corresponding equation [tex]\(4(x + 2)^2 = 0\)[/tex].

3. Simplify the equation:
[tex]\[ 4(x + 2)^2 = 0 \][/tex]
[tex]\[ (x + 2)^2 = 0 \][/tex]

4. Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]

5. Analyze the inequality:

The expression [tex]\((x+2)^2\)[/tex] represents a squared term, which is always non-negative. Therefore, [tex]\((x+2)^2\)[/tex] is zero when [tex]\(x = -2\)[/tex].

So, [tex]\(4(x + 2)^2\)[/tex] is zero when [tex]\(x = -2\)[/tex] and is positive for all other values of [tex]\(x\)[/tex].

6. Determine the inequality: Since [tex]\(4(x + 2)^2 \leq 0\)[/tex] implies that the expression is less than or equal to zero, this can only be true when [tex]\(4(x + 2)^2 = 0\)[/tex]. As calculated earlier, this happens only when [tex]\(x = -2\)[/tex].

7. Conclusion: The inequality [tex]\(4(x + 2)^2 \leq 0\)[/tex] holds true only when [tex]\(x = -2\)[/tex].

Therefore, the solution set for the quadratic inequality [tex]\(4(x + 2)^2 \leq 0\)[/tex] is:
[tex]\[ \{x \mid x = -2\} \][/tex]

Answer: [tex]\(\{x \mid x = -2\}\)[/tex]