Given the equation: $$x^2 - 4x + q = 0$$ 2.1.1 Determine the numerical value of the discriminant if $q = 4$ - NSC Technical Mathematics - Question 2 - 2023 - Paper 1

Since the discriminant is $0$, this indicates that the quadratic equation has exactly one real root (or a repeated real root). This means the roots are equal.

For the roots of the quadratic equation to be non-real, the discriminant must be less than $0$:

$riangle = b^2 - 4ac < 0$

Using $a = 1$, $b = -4$, and letting $c = p$, we have:

$riangle = (-4)^2 - 4(1)(p) < 0$

This simplifies to:

$16 - 4p < 0$

Solving for $p$ yields:

$16 < 4p$

Dividing both sides by $4$ gives:

$p > 4$

Thus, the numerical value(s) of $p$ for which the equation will have non-real roots is $p > 4$.