Difference between revisions of "1985 OIM Problems/Problem 3"
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== Problem == | == Problem == | ||
Find the roots <math>r_1</math>, <math>r_2</math>, <math>r_3</math>, and <math>r_4</math> of the equation: | Find the roots <math>r_1</math>, <math>r_2</math>, <math>r_3</math>, and <math>r_4</math> of the equation: | ||
| − | <cmath>4x^4-ax^3+bx^2- | + | <cmath>4x^4-ax^3+bx^2-cx+5=0</cmath> |
knowing that they're all real, positives and that: | knowing that they're all real, positives and that: | ||
| − | <cmath>frac{r_1}{2}+frac{r_2}{4}+frac{r_3}{5}+frac{r_4}{8}=1</cmath> | + | <cmath>\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1</cmath> |
| + | |||
| + | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== 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</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 == | ||
| + | https://www.oma.org.ar/enunciados/ibe1.htm | ||
Latest revision as of 22:47, 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, ![$\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\ge4\sqrt[4]{r_1r_2r_3r_4\frac{1}{5(64)}}=1$](http://latex.artofproblemsolving.com/f/a/9/fa9096f74534c51538c0231f5ccedfb060fa668e.png)
, so by the equality condition 
, so 
.
