Difference between revisions of "1985 OIM Problems/Problem 3"

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== Solution ==
 
== Solution ==
{{solution}}
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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 22:46, 8 April 2024

Problem

Find the roots $r_1$, $r_2$, $r_3$, and $r_4$ of the equation: \[4x^4-ax^3+bx^2-cx+5=0\] knowing that they're all real, positives and that: \[\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

By Vieta's, $r_1r_2r_3r_4=\frac54$. Because the roots are real and positive, by AM-GM, $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\ge4\sqrt[4]{r_1r_2r_3r_4\frac{1}{5(64)}}=1, so by the equality condition$\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14$, so$(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$.

See also

https://www.oma.org.ar/enunciados/ibe1.htm