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 10 - 2008 - Paper 1

Next, we will solve the inequality $q^2 + 8q < 0$.

To find the roots of the equation $q^2 + 8q = 0$, we factor it:

$q(q + 8) = 0$

The roots are:

• $q = 0$
• $q = -8$

These roots divide the number line into intervals: $(- ext{∞}, -8)$, $(-8, 0)$, and $(0, ext{∞})$.

Next, we test a point from each interval:

• For $q < -8$ (e.g., $q = -9$): $(-9)^2 + 8(-9) = 81 - 72 = 9$ (positive)
• For $-8 < q < 0$ (e.g., $q = -4$): $(-4)^2 + 8(-4) = 16 - 32 = -16$ (negative)
• For $q > 0$ (e.g., $q = 1$): $(1)^2 + 8(1) = 1 + 8 = 9$ (positive)

The inequality $q^2 + 8q < 0$ holds in the interval:

$-8 < q < 0$

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

$q ext{ is in } (-8, 0)$