[Cambridge/9794/Specimen Paper for 2010]

1. Solve the given quadratic equation to find its roots.

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)}$
$x = \frac{6 \pm \sqrt{36 - 52}}{2}$
$x = \frac{6 \pm \sqrt{-16}}{2}$
$x = \frac{6 \pm 4i}{2} = 3 \pm 2i$
The roots are $3 + 2i$ and $3 - 2i$.

2. Identify $p$ and $q$ based on the condition that $p$ has an acute argument.

For $3 + 2i$, the argument is $\tan^{-1}\left(\frac{2}{3}\right)$, which is an acute angle (since it is positive and less than $\frac{\pi}{2}$).
For $3 - 2i$, the argument is $\tan^{-1}\left(-\frac{2}{3}\right) = -\tan^{-1}\left(\frac{2}{3}\right)$, which is negative and not an acute angle.
Thus, $p = 3 + 2i$ and $q = 3 - 2i$.

3. Calculate the roots for the new equation, $ip$ and $-iq$.

The new roots are:
$ip = i(3 + 2i) = 3i + 2i^2 = -2 + 3i$
$-iq = -i(3 - 2i) = -3i + 2i^2 = -2 - 3i$

4. Find the sum and product of the new roots to form the new quadratic equation.

Sum of new roots $= (-2 + 3i) + (-2 - 3i) = -4$
Product of new roots $= (-2 + 3i)(-2 - 3i) = (-2)^2 - (3i)^2 = 4 - (-9) = 13$

The required equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$:
$x^2 - (-4)x + 13 = 0$
$x^2 + 4x + 13 = 0$