Given the roots: $x = \frac{-8 \pm \sqrt{q - 3}}{2}$ Describe the nature of the roots: 2.1.1 $q = 5$ 2.1.2 $q = 3$ 2.1.3 $q < 0$ 2.2 Determine for which value(s) of $p$ will the equation $3x^3 + 7x = 2x + p$ have non-real roots. - NSC Technical Mathematics - Question 2 - 2018 - Paper 1

For $q = 3$, we substitute again into the root expression:

$x = \frac{-8 \pm \sqrt{3 - 3}}{2} = \frac{-8 \pm \sqrt{0}}{2}$

Here, since the square root of zero is zero, both roots are equal. Therefore, the nature of the roots is equal.

When $q < 0$, note that the expression under the square root will be negative:

$q - 3 < 0 \Rightarrow \sqrt{q - 3} \text{ is imaginary}$

Therefore, since the square root yields an imaginary number, the roots are considered non-real.

First, we rearrange the equation into standard form:

$3x^3 + 7x - 2x - p = 0 \Rightarrow 3x^3 + 5x - p = 0$

For this cubic equation to have non-real roots, the discriminant must be less than zero. The discriminant ($D$) for a cubic equation can be complex, so a simplified approach is to investigate critical values.

By analyzing the derivative:
$f'(x) = 9x^2 + 5$
Since $f'(x) > 0$ for all $x$, the function is always increasing, leading to only one real root when there is a transition in the behavior at $p$.
Setting $p = 25/12$, a critical value for a negative effect leads to the inequality:
$(5)^2 - (4)(3)(-p) < 0 \Rightarrow 25 + 12p < 0 \Rightarrow p < \frac{-25}{12}$

Thus, the values of $p$ that will cause the equation to have non-real roots are given by the condition $p < \frac{-25}{12}$.