Given the roots: $x = 5 \, \pm \, \sqrt{k - 9}$ Describe the nature of the roots if: 2.1.1 $k < 9$ 2.1.2 $k = 9$ Determine the value(s) of $q$ for which the equation $-x^2 + 2qx - 4 = 0$ will have non-real roots. - NSC Technical Mathematics - Question 2 - 2021 - Paper 1

Substituting $k = 9$ into the root expression gives:

$x = 5 \, \pm \, \sqrt{9 - 9} = 5 \, \pm \, 0$.

This results in a single repeated root, which is real, rational, and equal.

Thus, when $k = 9$, the roots are real, rational, and equal.

To find the values of $q$ that result in non-real roots, we calculate the discriminant:

$\Delta = b^2 - 4ac$

For our equation, $a = -1$, $b = 2q$, and $c = -4$:

$\Delta = (2q)^2 - 4(-1)(-4) = 4q^2 - 16$

Setting the discriminant less than zero for non-real roots gives:

$4q^2 - 16 < 0$

Rearranging leads to:

$4q^2 < 16$

Dividing by 4:

$q^2 < 4$

Taking the square root results in:

$-2 < q < 2$

Thus, the values of $q$ for which the equation has non-real roots are in the interval $(-2, 2)$.