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Difference between revisions of "2018 AIME II Problems/Problem 6"

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==Solution==
 
==Solution==
  
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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.
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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:
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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.
  
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We should set the discriminant of the quadratic greater than or equal to 0.
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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>.
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<math>18 + 19 = \boxed{037}</math>
 
<math>18 + 19 = \boxed{037}</math>
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 +
~First
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==Video Solution==
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https://www.youtube.com/watch?v=q2oc7n-n6aA
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~Shreyas S
  
 
==See Also:==
 
==See Also:==
  
 
{{AIME box|year=2018|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2018|n=II|num-b=5|num-a=7}}
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[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:45, 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)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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