<p>Solve the inequality: <span>\((k-2)x^2 + 8x + (k+4) > 0\)</span> for all \(x \in \mathbb{R}\)</p><p>Find the least integral value of \(k\).</p>
Step-by-Step Solution
Key Concept: For a quadratic to be positive for all real values, its discriminant must be non-positive (ensuring no real roots and parabola opens upward).
<p><strong>Step 1:</strong> For a quadratic inequality \((k-2)x^2 + 8x + (k+4) > 0\) to hold for all \(x \in \mathbb{R}\), the discriminant must be non-positive.</p><p><strong>Step 2:</strong> Calculate discriminant: \(\Delta = 64 - 4(k-2)(k+4) \geq 0\)</p><p><strong>Step 3:</strong> Simplify: \(16 - (k-2)(k+4) \geq 0\)</p><p><strong>Step 4:</strong> \((k-2)(k+4) \leq 16\)</p><p><strong>Step 5:</strong> \(k^2 + 2k - 8 \leq 16\)</p><p><strong>Step 6:</strong> \(k^2 + 2k - 24 \leq 0\)</p><p><strong>Step 7:</strong> \((k+6)(k-4) \leq 0\)</p><p><strong>Step 8:</strong> \(-6 \leq k \leq 4\)</p><p>∴ The least integral value of \(k\) is \(-6\).</p>
Correct Answer: -6