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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 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 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 becomes:
 
The polynomial becomes:
Line 29: Line 29:
 
<math>a \leq -\dfrac{1}{2}</math>
 
<math>a \leq -\dfrac{1}{2}</math>
  
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This means that the interval <math>(-\dfrac{1}{2}, \dfrac{3}{2})</math> is the "bad" interval. The length of the interval where <math>a</math> can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the good interval is 36 units long.
+
This means that the interval <math>\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)</math> is the "bad" interval. The length of the interval where <math>a</math> can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.
  
 
<math>\dfrac{36}{38} = \dfrac{18}{19}</math>
 
<math>\dfrac{36}{38} = \dfrac{18}{19}</math>
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<math>18 + 19 = \boxed{037}</math>
 
<math>18 + 19 = \boxed{037}</math>
  
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==See Also==
+
~First
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=q2oc7n-n6aA
 +
~Shreyas S
 +
 
 +
==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}}
 +
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:20, 27 March 2022

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

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

See Also:

2018 AIME II (Problems • Answer Key • Resources)
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