Explanation:
Let r be the common root of the given equations.
Then we have:
ar³ + 4r² - 5r - 10 = 0 ...(1)
and
ar³ - 9r - 2 = 0 ... (2)
Subtracting equation (2) from equation (1), we get:
13r² - 3r - 8a = 0
Solving for r using quadratic formula, we get:
r = [3 ± sqrt(9 + 52a)]/26
For r to be a real number, the discriminant of the quadratic expression inside the square root must be non-negative:
9 + 52a ≥ 0
Solving for a, we get:
a ≥ -9/52
Therefore, the possible values of a are all real numbers greater than or equal to -9/52.