2016 AMC 12A Problems/Problem 24: Difference between revisions
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<cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath> | <cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath> | ||
<cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{( | <cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}</cmath> | ||
The actual function: | The actual function: | ||
Revision as of 20:31, 5 February 2016
Problem
There is a smallest positive real number 
such that there exists a positive real number 
such that all the roots of the polynomial 
are real. In fact, for this value of the value of is unique. What is this value of ?
Solution
The acceleration must be zero at the 
-intercept; this intercept must be an inflection point for the minimum value.
Derive 
so that the acceleration 
:
for the inflection point/root.
The function with the minimum :
![\[f(x)=\left(x-\frac{a}{3}\right)^3\]](http://latex.artofproblemsolving.com/b/9/4/b94a95664c3844462897054d5ee0063e598ecd94.png)
![\[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\]](http://latex.artofproblemsolving.com/0/3/9/03996f81ab32767c05458c60e84b59c2f6be2b7e.png)
Since this is equal to the original equation ,
![\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\]](http://latex.artofproblemsolving.com/1/b/0/1b0f1b53e7fb633d1fd2778ca4fa652a5d1df6f4.png)
![\[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}\]](http://latex.artofproblemsolving.com/f/8/5/f85227f32d17658231598032a4092f0ea353fffd.png)
The actual function:
triple root! "Complete the cube."
See Also
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America. 
