Question

what is the solution set of the quadratic inequality $4(x + 2)^2 \leq 0$?
$\\{x| x = -2\\}$
$\\{x| x \in \mathbb{r}\\}$
$\\{x| x = 2\\}$
$\varnothing$

Explanation:

Step1: Analyze the inequality

We have the inequality \(4(x + 2)^2\leq0\). First, note that the coefficient \(4\) is positive, and \((x + 2)^2\) is a square of a real number, so \((x + 2)^2\geq0\) for all real numbers \(x\).

Step2: Solve the inequality

Since \(4>0\) and \((x + 2)^2\geq0\), the product \(4(x + 2)^2\geq0\) for all real \(x\). We want to find when \(4(x + 2)^2\leq0\). The only way the product of a positive number (\(4\)) and a non - negative number \((x + 2)^2\) is less than or equal to \(0\) is when both conditions are met: \(4>0\) (always true) and \((x + 2)^2 = 0\) (because if \((x + 2)^2>0\), then \(4(x + 2)^2>0\)).
Solving \((x + 2)^2=0\), we take the square root of both sides: \(x+2 = 0\), so \(x=-2\). So the solution set is \(\{x|x = - 2\}\) (assuming one of the options is \(\{x|x=-2\}\), maybe a typo in the original options, but from the analysis, when \(x = - 2\), \(4(-2 + 2)^2=4\times0 = 0\) which satisfies \(4(x + 2)^2\leq0\), and for any other \(x\), \((x + 2)^2>0\) so \(4(x + 2)^2>0\) which does not satisfy the inequality).

Answer:

\(\{x|x = - 2\}\) (assuming the option with \(x=-2\) is the correct one, likely a formatting issue in the original options, but based on the solution of the inequality \(4(x + 2)^2\leq0\), the solution is \(x=-2\))