(-16) : (((х+2) ²)-5) ≥ 0

To solve the inequality \((-16) : (((x+2)²)-5) ≥ 0\), we start by rewriting it in a more manageable form. The expression \(-16\) divided by another expression is greater than or equal to zero. First, let's break down the inequality: \[ -\frac{16}{((x+2)² - 5)} \geq 0 \] Multiplying both sides of the inequality by \(-1\) (remembering to flip the inequality sign) gives us: \[ \frac{16}{((x+2)² - 5)} \leq 0 \] Now, we need to determine when the fraction \(\frac{16}{((x+2)² - 5)}\) is less than or equal to zero. This fraction will be non-positive (i.e., less than or equal to zero) when the numerator is positive (which it always is since \(16\) is positive) and the denominator \(((x+2)² - 5)\) is negative or zero. Now, let's find the values of \(x\) that make the denominator \(((x+2)² - 5)\) equal to zero: \[ (x+2)² - 5 = 0 \] Solving this gives: \[ (x+2)² = 5 \] Taking square roots: \[ x + 2 = \pm \sqrt{5} \] Thus: \[ x = -2 + \sqrt{5} \quad \text{or} \quad x = -2 - \sqrt{5} \] Now we need to check when \((x+2)² - 5 < 0\). The critical points obtained are \(x = -2 + \sqrt{5}\) and \(x = -2 - \sqrt{5}\). To analyze the sign of the denominator in intervals defined by these critical points, we check intervals: 1. \(x < -2 - \sqrt{5}\) 2. \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\) 3. \(x > -2 + \sqrt{5}\) **Checking intervals:** 1. For \(x < -2 - \sqrt{5}\): Choose \(x = -10\): \((x+2) = -8\) so \((x + 2)^2 = 64 - 5 = 59\) (positive). 2. For \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\): Choose \(x = -2\): \((x + 2) = 0\) so \((x + 2)^2 = 0 - 5 = -5\) (negative). 3. For \(x > -2 + \sqrt{5}\): Choose \(x = 2\): \((x + 2) = 4\) so \((x + 2)^2 = 16 - 5 = 11\) (positive). **Conclusion:** The denominator \((x + 2)^{2} - 5 \leq 0\) when \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\). At the endpoints: - At \(x = -2 + \sqrt{5}\), the denominator equals \(0\). - At \(x = -2 - \sqrt{5}\), the denominator also equals \(0\). Thus, the solution set for the original inequality \((-16) : (((x+2)²)-5) ≥ 0\) is: \[ x \in [-2 - \sqrt{5}, -2 + \sqrt{5}] \] Or, in interval notation: \[ [-2 - \sqrt{5}, -2 + \sqrt{5}] \]