# 2018 AIME II Problems/Problem 6

## Problem

A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$. The probability that the roots of the polynomial $x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$

are all real can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

## Solution

The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.

The polynomial becomes: $(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$

Since we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.

We should set the discriminant of the quadratic greater than or equal to 0. $(2a - 1)^2 - 4 \geq 0$.

This simplifies to: $a \geq \dfrac{3}{2}$

or $a \leq -\dfrac{1}{2}$

This means that the interval $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$ is the "bad" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long. $\dfrac{36}{38} = \dfrac{18}{19}$ $18 + 19 = \boxed{037}$

~First

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 