## Problem

What is the probability that the following quadratic equation has real roots? $x^2 + 2bx + c =0$

## Solution

We begin by applying the quadratic formula to the equation given above:

$x = \frac{-2b \pm \sqrt{4b^2-4c}}{2}$

Rearranging the contents of $$\sqrt{4b^2-4c}$$ shows that the roots are real if and only if the following holds:

$4b^2-4c > 0$ $c < b^2$

We can calculate the probability that $$c < b^2$$ geometrically. We begin by drawing a $$2n$$ by $$2n$$ square centered at the origin. The graph of $$c=b^2$$ intersects the square at $$b=\pm \sqrt{n}$$. Within the square, the shaded region below the parabola represents equations with real roots.
To find the probability of real roots, $$P$$, we can calculate the proportion of the shaded area with respect to the entire square. We will consider the limiting value of this proportion as $$n$$ tends towards $$\infty$$:
$\begin{split} P &= \lim_{n \to \infty} \frac{4n^2 - \int_{-\sqrt{n}}^{\sqrt{n}}n-b^2db}{4n^2} \\ &= \lim_{n \to \infty} \frac{4n^2 - \frac{4}{3}n^{\frac{3}{2}}}{4n^2} \\ &= \lim_{n \to \infty} 1 - \frac{3}{\sqrt{n}} \\ &= 1 \end{split}$
Thus, the probability of the equation having real roots is $$1$$.