Find The Probability of Quadratic having imaginary roots Two numbers$\ p$ and$\ q$ are both chosen randomly (and independently of each other) from the interval$\ [-2, 2]$. Find the probability that$\ 4x^2+4px+1-q^2=0$ has imaginary roots.
How do you solve this problem? Given that we're trying to find out when the quadratic has imaginary roots I suppose we use the discriminant. Which, $\ b^2-4ac=(4p)^2-4(4)(1-q^2)$. It can then be factored out to$\ 4^2(p^2-(1-q^2)=4^2(p^2-(1-q)(1+q))$ 
I'm not even sure if I'm on the right track here. The answer given to us was $\ \frac{\pi}{16}$ and I'm still at a lost on how to get there. An explanation would be appreciated.
 A: Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 \lt 1$ which corresponds to a circle of area $\pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.
A: If there are two imaginary roots to a quadratic equation, then the discriminant is negative.
$$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
$$p^2+q^2<1$$
The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$.  The circle is a circle of radius 1, and therefore has an area of $\pi$.  The rectangle is a square of side length 4, and therefore has an area of 16.  The ratio is therefore $\frac{\pi}{16}$
A: To nitpick:
I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$
For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.
A: Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$\left\{(p,q)\in[-2,2]\times[-2,2]\,\middle|\,D(p,q)<0\right\},$$then$$\frac A{16}=\frac\pi4,$$where that $16$ is the area of the square $[-2,2]\times[-2,2]$.
