Given that the equation $2qx^2 + qx - 1 = 0$, where $q$ is a constant, has no real roots, (a) show that $q^2 + 8q < 0$ - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 1

Next, we can factor the inequality $q^2 + 8q < 0$:

$q(q + 8) < 0$

To solve this, we identify the critical points by setting the expression to zero:

1. $q = 0$
2. $q + 8 = 0 ightarrow q = -8$

We now test intervals around these critical points:

1. For $q < -8$, both factors are negative, so the product is positive.
2. For $-8 < q < 0$, $q$ is negative and $(q + 8)$ is positive, making the product negative (this is the interval we want).
3. For $q > 0$, both factors are positive, so the product is positive.

Thus, the solution set is the interval:

$-8 < q < 0$

Hence, the set of possible values of $q$ is:

$(-8, 0)$.