Difference between revisions of "1985 OIM Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | By Vieta's, <math>r_1r_2r_3r_4=\frac54</math>. Because the roots are real and positive, by AM-GM, <math>\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\ge4\sqrt[4]{r_1r_2r_3r_4\frac{1}{5(64)}}=1, so by the equality condition </math>\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14<math>, so </math>(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$. |
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe1.htm | https://www.oma.org.ar/enunciados/ibe1.htm | ||
Revision as of 23:46, 8 April 2024
Problem
Find the roots 
, 
, 
, and 
of the equation:
![\[4x^4-ax^3+bx^2-cx+5=0\]](http://latex.artofproblemsolving.com/d/9/3/d93275ae96fdc4e2a22e121aab01b3fc03049a4e.png)
knowing that they're all real, positives and that:
![\[\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1\]](http://latex.artofproblemsolving.com/d/1/9/d195dc3007c6d7345fb00b1bdacae607e880987d.png)
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
By Vieta's, 
. Because the roots are real and positive, by AM-GM, ![$\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\ge4\sqrt[4]{r_1r_2r_3r_4\frac{1}{5(64)}}=1, so by the equality condition$](http://latex.artofproblemsolving.com/5/d/3/5d3019a969257140ddb0590d9b0a584d98c862fb.png)
\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14
(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$.
