Let's solve the inequality given in the problem: \(\frac{x + 1}{x^2 + 2} > \frac{1}{4}\). We want to find the number of integral solutions for this inequality.
First, we'll set the inequality by cross-multiplying to clear the fractions:
\(4(x + 1) > x^2 + 2\)
Expand and simplify the inequality:
\(4x + 4 > x^2 + 2\)
\(4x + 4 - 2 > x^2\)
\(4x + 2 > x^2\)
\(-x^2 + 4x + 2 > 0\) or equivalently \(x^2 - 4x - 2 < 0\).
Find the roots of the quadratic equation \(x^2 - 4x - 2 = 0\) using the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -2\):
Discriminant: \(b^2 - 4ac = 16 + 8 = 24\)
Roots: \(x = \frac{{4 \pm \sqrt{24}}}{2} = \frac{{4 \pm 2\sqrt{6}}}{2} = 2 \pm \sqrt{6}\)
Identify the range where the quadratic inequality is true:
Since \(x^2 - 4x - 2 < 0\) is a parabola opening upwards, solutions for the inequality are between the roots \(2 - \sqrt{6} < x < 2 + \sqrt{6}\).
Numerically, \(2 - \sqrt{6} \approx -0.449\) and \(2 + \sqrt{6} \approx 4.449\).
Determine integer solutions within this range:
Integer values of \(x\) within this range are 0, 1, 2, 3, 4, making 5 solutions.
Thus, the number of integral solutions is 5. This matches the correct answer.