Question

what is the solution set of the quadratic inequality 4(x + 2)² ≤ 0?
{ x | x = -2 }
{ x | x ∈ ℝ }
{ x | x = 2 }
∅

Explanation:

Step1: Analyze the square term

A square of any real number \(a\), i.e., \(a^{2}\), is always non - negative. That is, for any real number \(x\), \((x + 2)^{2}\geq0\).

Step2: Analyze the coefficient and the inequality

We have the inequality \(4(x + 2)^{2}\leq0\). Since \(4>0\), when we multiply a non - negative number \((x + 2)^{2}\) by a positive number \(4\), the result \(4(x + 2)^{2}\) is also non - negative. So \(4(x + 2)^{2}\geq0\) for all real numbers \(x\).
The only way for \(4(x + 2)^{2}\leq0\) to hold is when \(4(x + 2)^{2}=0\) (because it can't be negative).
Set \(4(x + 2)^{2}=0\). Divide both sides by \(4\), we get \((x + 2)^{2}=0\). Taking the square root of both sides, \(x+2 = 0\), so \(x=-2\).

Answer:

\(\{x|\ x = - 2\}\)