Question:

<p><strong>Step 1:</strong> Rewrite the equation by factoring out x from numerators and manipulating denominators:</p><p>Divide each fraction's numerator and denominator strategically. Note that x² + x + 3 and x² - 5x + 3 can be related through substitution.</p><p><strong>Step 2:</strong> Rewrite fractions as:</p><p>$$\frac{4}{x + 1 + \frac{3}{x}} + \frac{5}{x - 5 + \frac{3}{x}} = -\frac{3}{2}$$</p><p><strong>Step 3:</strong> Let $y = x + \frac{3}{x}$. Then:</p><p>$$\frac{4}{y + 1} + \frac{5}{y - 5} = -\frac{3}{2}$$</p><p><strong>Step 4:</strong> Clear denominators:</p><p>$$4(y-5) + 5(y+1) = -\frac{3}{2}(y+1)(y-5)$$</p><p>$$4y - 20 + 5y + 5 = -\frac{3}{2}(y^2 - 4y - 5)$$</p><p>$$9y - 15 = -\frac{3}{2}y^2 + 6y + \frac{15}{2}$$</p><p><strong>Step 5:</strong> Simplify to get:</p><p>$$3y^2 + 6y - 45 = 0$$</p><p>$$y^2 + 2y - 15 = 0$$</p><p>$$(y + 5)(y - 3) = 0$$</p><p>So $y = -5$ or $y = 3$</p><p><strong>Step 6:</strong> For $y = 3$: $x + \frac{3}{x} = 3$ → $x^2 - 3x + 3 = 0$ → $\Delta = 9 - 12 = -3 < 0$ (no real roots)</p><p><strong>Step 7:</strong> For $y = -5$: $x + \frac{3}{x} = -5$ → $x^2 + 5x + 3 = 0$ → $\Delta = 25 - 12 = 13 > 0$ (two distinct real roots, both irrational since $\sqrt{13}$ is irrational)</p><p>∴ Answer: <strong>A</strong> (The number of irrational roots is <strong>2</strong>)