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2018 AIME II Problems/Problem 6: Difference between revisions

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==Solution==
==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, <math>x = 1, -1, 2, -2</math> are all possible rational roots. Upon plugging these roots into the polynomial, <math>x = -2</math> and <math>x = 1</math> make the polynomial equal 0 and thus, they are roots that we can factor out.
The polynomial we are given is rather complicated, so let's use [[Rational Root Theorem]] to see if it has any retinal roots. By Rational Root Theorem, <math>x = 1, -1, 2, -2</math> are all possible rational roots. Upon plugging these roots into the polynomial, <math>x = -2</math> and <math>x = 1</math> make the polynomial equal 0 and thus, they are roots that we can factor out.


The polynomial becomes:
The polynomial becomes:
Line 17: Line 17:
Since we know <math>1</math> and <math>-2</math> are real numbers, we only need to focus on the quadratic.
Since we know <math>1</math> and <math>-2</math> 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.
Set the discriminant of the quadratic greater than or equal to 0, to ensure the remaining roots are real.


<math>(2a - 1)^2 - 4 \geq 0</math>.
<math>(2a - 1)^2 - 4 \geq 0</math>.
Line 34: Line 34:


<math>18 + 19 = \boxed{037}</math>
<math>18 + 19 = \boxed{037}</math>
~First


==Video Solution==
==Video Solution==

Latest revision as of 15:45, 22 July 2025

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 let's use Rational Root Theorem to see if it has any retinal roots. By 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.

Set the discriminant of the quadratic greater than or equal to 0, to ensure the remaining roots are real.

$(2a - 1)^2 - 4 \geq 0$.

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

https://www.youtube.com/watch?v=q2oc7n-n6aA ~Shreyas S

See Also:

2018 AIME II (Problems • Answer Key • Resources)
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