# 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.

• Hint: $p^2-(1-q^2) < 0 \Rightarrow p^2 + q^2 < 1$ – gandalf61 Dec 3 '18 at 15:00
• As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0. – NickD Dec 3 '18 at 15:30

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 $$\frac{\pi}{16}$$

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.

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.

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 $$[-2,2]\times[-2,2]$$.