Final answer:
All values of 'a' less than -3/8 will result in the quadratic equation ax^2 + 3x - 6 having two imaginary roots, as the discriminant will be negative for these values.
Step-by-step explanation:
To find all the values of 'a' for which the quadratic equation ax^2 + 3x - 6 has two imaginary roots, we need to use the discriminant part of the quadratic formula, which is b^2 - 4ac. For a quadratic equation to have imaginary roots, the discriminant must be negative (less than zero).
The quadratic equation in question is in the standard form ax^2 + bx + c, where a is the coefficient of x^2, b is the coefficient of x, and c is the constant term. In this case, b = 3 and c = -6.
The discriminant is therefore given by (3)^2 - 4(a)(-6). For two imaginary roots, we require:
(3)^2 - 4(a)(-6) < 0
Solving this inequality:
9 + 24a < 0
24a < -9
a < -\(rac{9}{24}\)
a < -\(rac{3}{8}\)
Hence, the quadratic equation ax^2 + 3x - 6 has two imaginary roots for all values of a that are less than -\(rac{3}{8}\).