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

m (Solution)
 
Line 8: Line 8:
  
 
== 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)}$.
+
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</math>, 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)}</math>.
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe1.htm
 
https://www.oma.org.ar/enunciados/ibe1.htm

Latest revision as of 22:47, 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